Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 195 5. Find the locus of a point P(x, y) such that the triangle with the points P(x, y), A(5, 0) and the origin, O, as vertices is right-angled at P. 6. Find the cartesian equations of the loci of points with the following parametric equations. (a) x = t 2 – 1, y = 3 + t (b) x = 3t 2 , y = 2t 3 (c) x = 3t + 1, y = 3 t – 1 (d) x = 1 + t t , y = 1 – t t 2 (e) x = 6t, y = 3t 2 + 1 (f) x = t 2 – 1, y = t 3 + 1 (g) x = 1 + 2t, y = 2 t 2 (h) x = 4t, y = 4 t 2 – 1 Conic sections (conics) A conic section (or just conic) is a curve obtained as the intersection of a cone (more precisely, a right circular conical surface) with a plane (see Figure 5.1). In Analytic Geometry, a conic may be defined as a plane algebraic curve of degree 2. Traditionally, the three types of conic sections are the hyperbola, the parabola, and the ellipse. The circle is a special case of the ellipse, and it is sometimes called the fourth type of conic section. A conic can also be defined as the locus of a point satisfying certain conditions. Circle Ellipse Parabola Hyperbola Figure 5.1 The general equation for a conic is Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 The parabola A parabola is the locus of a point whose distance from a fixed point, called the focus, is always equal to its perpendicular distance from a fixed straight line, called the directrix. The point of intersection of the parabola and its axis of symmetry is called the vertex. In the diagram in Figure 5.2, let S(a, 0) be the fixed point and x = –a be the fixed line. Let P(x, y) be the moving point. By definition, PS = PN PS2 = PN2 (x – a) 2 + (y – 0)2 = (x + a) 2 x2 – 2ax + a2 + y2 = x2 + 2ax + a2 y2 = 4ax This is the standard form of the equation of a parabola with its vertex at the origin. –a O S (a, 0) x = –a N P (x, y) y x y 2 = 4ax Figure 5.2 The Beauty of Conics INFO
196 Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 The distance between the vertex and the focus is the focal length. Since the power of y is even, the graph is symmetrical about the x-axis and exists only for x 0. O y N a x x = h – a S(h + a, k) (h, k) P(x, y) Figure 5.3 Suppose that the parabola with focal length a has its vertex at (h, k). Then the focus is at (h + a, k) and its directrix is at x = h – a. Let P(x, y) be a point on the parabola. By definition, PS = PN. Therefore, the locus of P is defined by [x – (h + a)]2 + (y – k) 2 = [x – (h – a)]2 x2 – 2(h + a)x + (h + a) 2 + (y – k) 2 = x2 – 2(h – a)x + (h – a) 2 (y – k) 2 = 4a(x – h) For example, consider the equation (y – 2)2 = 8(x – 3) = 4(2)(x – 3) It represents a parabola with vertex at the point (3, 2) and has focal length 2. The focus is at (5, 2) and directrix is x = 1. Example 4 Sketch the curve y2 = 2x. Show that y2 = 2(x + 2y) is the equation of a parabola and state the coordinates of its vertex. Sketch this curve. Solution: The equation y2 = 2x can be written as y2 = 41 1 2 2x. Hence, it is a parabola with vertex at O, and is symmetrical about the x-axis. The graph of y2 = 2x is as shown below. y 2 = 2x y x O y 2 = 2(x + 2y) y = 2 Y y X x O (–2, 2) For y2 = 2(x + 2y) y2 – 4y = 2x Completing the square for the LHS, y2 – 4y + 4 = 2x + 4 (y – 2)2 = 2(x + 2) = 41 1 2 2(x + 2) This is the equation of a parabola with vertex at (–2, 2) and focal length 1 2 . Its focus is at 1– 3 2 , 22 and directrix is at x = – 5 2 .
Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 197 Parametric equations A parabola can also be defined by parametric equations. Let x = at2 and y = 2at, where t is the parameter. Then, y2 = 4a2 t 2 = 4a(at 2 ) = 4ax Hence, the equation of the locus of the point P can be written as x = at2 , y = 2at, t R Any point lying on the parabola y2 = 4ax can be represented by the parametric coodinates (at2 , 2at). Example 5 Find the loci of the following points. (a) P(2t 2 , 4t) (b) Q1 1 2 t 2 , t2 Solution: (a) Let x = 2t 2 and y = 4t. Squaring y = 4t ⇒ y2 = 16t 2 = 8(2t 2 ) = 8x Hence, the locus of P is the parabola y2 = 8x. (b) Let x = 1 2 t 2 and y = t y2 = t 2 = 21 1 2 t 2 2 = 2x Hence, the locus of Q is the parabola y2 = 2x. Example 6 For each of the following curves, obtain its parametric equations. (a) y2 = 12x (b) 2y2 = 9x Solution: (a) Rewriting y2 = 12x as y2 = 4(3)x Comparing with the standard equation y2 = 4ax, we see that a = 3. For y2 = 4ax, its parametric equations are x = at2 , y = 2at. Hence, for y2 = 12x, its parametric equations are x = 3t 2 , y = 6t. (b) Rewriting 2y2 = 9x as y2 = 9 2 x y2 = 41 9 8 2x Comparing with y2 = 4ax, we see that a = 9 8 . Hence, for 2y2 = 9x, its parametric equations are x = 9 8 t 2 , y = 9 4 t.
198 Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 Chords, tangents and normals to a parabola Suppose P(ap2 , 2ap) and Q(aq2 , 2aq) are two points on the parabola y2 = 4ax. Then, gradient of PQ = 2ap – 2aq ap2 – aq2 = 2a(p – q) a(p – q)(p + q) = 2 p + q Hence, the gradient of the chord joining the points P(ap2 , 2ap) and Q(aq2 , 2aq) is 2 p + q . Equation of the chord PQ is y – 2ap = 2 p + q (x – ap2 ) i.e. (p + q)y – 2ap(p + q) = 2x – 2ap2 (p + q)y – 2ap2 – 2apq = 2x – 2ap2 or (p + q)y = 2x + 2apq Now, as Q moves towards P, the chord PQ gets shorter, until, when q = p, the point Q coincides with the point P. The chord PQ becomes the tangent at P. Hence, gradient of tangent at P = 2 2p = 1 p Equation of tangent at P is 2py = 2x + 2ap2 i.e. py = x + ap2 Gradient of normal at P = –p Equation of normal at P is y – 2ap = –p(x – ap2 ) y + px = 2ap + ap3 Example 7 Obtain the equation of the chord joining the point P(ap2 , 2ap) and the point Q(aq2 , 2aq) on the parabola y2 = 4ax, and find the point of intersection of the tangents at P and Q. If PQ passes through (a, a), find the locus of the point of intersection of the two tangents. Solution: Equation of the chord PQ is (p + q)y = 2x + 2apq Equation of tangent at P is py = x + ap2 Equation of tangent at Q is qy = x + aq2 Tangent at P : py = x + ap2 …………… Tangent at Q: qy = x + aq2 …………… – : (q – p)y = a(q2 – p2 ) = a(q – p)(q + p) y = a(p + q) O y x P(ap2, 2ap) Q(aq2, 2aq) tangent at P Figure 5.4
Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 199 Substitute into , ap(p + q) = x + ap2 ap2 + apq = x + ap2 x = apq Hence, the point of intersection of the tangents at P and Q is [apq, a(p + q)]. Let (, b) ≡ [apq, a(p + q)] Equation of PQ : (p + q)y = 2x + 2apq If PQ passes through (a, a), then (p + q)a = 2a + 2apq i.e. b = 2a + 2. Hence, the locus of the point of intersection of the tangents at P and Q is y = 2a + 2x, a straight line. Example 8 PQ is a chord of the parabola x = at2 and y = 2at, and which passes through the focus (a, 0). Show that the locus of the mid-point of PQ is another parabola and hence, find its equation. Solution: Let the gradient of PQ be m. Equation of PQ which passes through (a, 0) is y – 0 = m(x – a) y = m(x – a) If (at2 , 2at) is a point on the parabola lying on PQ, then 2at = m(at2 – a) 2t = m(t 2 – 1) mt2 – 2t – m = 0 …………… ❊ Let t1 be the parameter for P, t2 the parameter for Q. Then t1 and t2 are the roots of the equation ❊ . Sum of the roots, t1 + t2 = 2 m Product of the roots, t1 t2 = –1 Let (, b) be the mid-point of PQ. a = 1 2 (at1 2 + at2 2 ) = 1 2 a[(t1 + t2 ) 2 – 2t1 t2 ] = 1 2 a1 4 m2 + 22 = a m2 (2 + m2 ) ………… b = 1 2 (2at1 + 2at2 ) = a(t1 + t2 ) = 2a m ……………
200 Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 From , m = 2a b m2 = 4a2 b2 Substitute into , = a 4a2 b2 12 + 4a2 b2 2 = b2 4a 12 + 4a2 b2 2 4a = 2b2 + 4a2 4a – 4a2 = 2b2 b2 = 2a(a – a) Hence, as PQ varies, the locus of the mid-point of PQ is y2 = 2a(x – a), which is another parabola. Exercise 5.2 1. Sketch the curve represented by each of the following equations. (a) y2 = 8x (b) y2 + 12x = 0 (c) 2y2 = 9x (d) 3y2 + 8x = 0 2. Find the vertex of each of the following parabolas and sketch its curve. (a) y2 = 4(x – 3) (b) (y + 2)2 = 2(x – 1) (c) y2 – 2y = 8x + 7 (d) y2 = 6y + x – 5 3. Find, in cartesian form, each of the following parametric equations. (a) x = 2t 2 , y = 4t (b) x = 3 2 t 2 , y = 3t (c) x = 3p2 , y = 6p (d) x = 5 3 t 2 , y = 10 3 t 4. Obtain the parametric equations for each of the following curves. (a) y2 = 6x (b) y2 = 9x (c) 2(y – 1)2 = 15x (d) 3y2 = 8(x – 1) (e) (y – 1)2 = 16(x – 2) 5. Find the equations of the tangents and normals for each of the following parabolas at the given points. (a) y2 = 8x; (2, 4) (b) y2 = 12x; (3, 6) (c) y2 = 10x; 1 5 2 , –52 (d) y2 = 6ax; 1 3 2 a, 3a2 6. The points (1, 6) and (81, –54) lie on the parabola y2 = 36x. Find the equation of the tangents to the parabola at these points, and hence show that the tangents are perpendicular to each other. Find the point of intersection of the tangents. 7. Find the equations of the tangent and normal to the parabola y2 = 4x at the point (t 2 , 2t). If P is the point with t = 2 , find the coordinates of the point Q where the normal at the point P meets the parabola again. If O is the origin, show that OP is perpendicular to OQ.
Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 201 8. If PQ is a chord which is a normal at the point (a, 2a), which lies on the parabola y2 = 4ax, find the length of PQ. Find also the area of the triangle formed by the chord and the tangents at P and Q. 9. The line y – 2x + 4a = 0 intersects the parabola y2 = 4ax at the point P(ap2 , 2ap) and the point Q(aq2 , 2aq). Find the value of p + q and pq. Hence, find the coordinates of the mid-point of PQ. 10. P(ap2 , 2ap) and Q(aq2 , 2aq) are two points on the parabola y2 = 4ax and the line PQ passes through the point (a, 0). Show that if the tangents at P and Q meet at T, the locus of T is a straight line and find its equation. 11. If the normal at P(ap2 , 2ap) to the parabola y 2 = 4ax meets the curve again at Q(aq2 , 2aq), show that p2 + pq + 2 = 0. Show that the equation of the locus of the point of intersection of the tangents at P and Q to the parabola is y2 (x + 2a) + 4a3 = 0. 12. The line y = mx + c intersects the parabola y2 = 4ax at the points P and Q. Show that the coordinates of the mid-point of PQ is 1 2a – mc m2 , 2a m 2. If this mid-point is M, find the locus of M when m varies and c = 1. The ellipse An ellipse is the locus of a point the sum of whose distances from two fixed points, called the foci, is constant. The mid-point of the line segment connecting the foci is called the centre. Let S and S' be the two fixed points on the x-axis with coordinates (c, 0) and (–c, 0) respectively. Hence, the origin O is the centre of the ellipse. If P(x, y) is any point on the ellipse, then the sum of the distances is given by PS + PS' = (x – c) 2 + (y – 0)2 + [x – (–c)]2 + (y – 0)2 = constant Let us suppose that this constant is 2a, with a . c. Then, (x – c) 2 + (y – 0)2 + [x – (–c)]2 + (y – 0)2 = 2a (x – c) 2 + y2 + (x + c) 2 + y2 = 2a. y x P(x, y) S'(–c, 0) O S(c, 0) Figure 5.5 Let A, A' be two points on the x-axis with coordinates (a, 0) and (–a, 0) respectively, where a . c. We see that AS + AS' = A'S + A'S' = AA' = 2a. Hence, A(a, 0) and A'(–a, 0) are points on the ellipse. Let B(b, 0), B'(–b, 0) be two points on the y-axis such that BS = BS' = a.
202 Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 Since BS + BS' = 2a, B and B' are also points on the ellipse, with a2 = b2 + c 2 , as shown in Figure 5.6 below. y x B(0, b) A'(–a, 0) S' O S c A(a, 0) B'(0, –b) b a Figure 5.6 From above, we have (x – c) 2 + y2 + (x + c) 2 + y2 = 2a (x + c) 2 + y2 = 2a – (x – c) 2 + y2 Squaring both sides, we have (x + c) 2 + y2 = 4a2 + (x – c) 2 + y2 – 4a (x – c) 2 + y2 x2 + 2cx + c 2 + y2 = 4a2 + x2 – 2cx + c 2 + y2 – 4a(x – c) 2 + y2 4a(x – c) 2 + y2 = 4a2 – 4cx a(x – c) 2 + y2 = a2 – cx Squaring both sides, we have a2 [(x – c) 2 + y2 ] = (a2 – cx) 2 a2 (x2 – 2cx + c 2 + y2 ) = a4 – 2a2 cx + c 2 x2 a2 x2 – 2a2 cx + a2 c 2 + a2 y2 = a4 – 2a2 cx + c 2 x2 x2 (a2 – c 2 ) + a2 y2 = a2 (a2 – c 2 ) Since a2 = b2 + c 2 , a2 – c 2 = b2 . Hence, b2 x2 + a2 y2 = a2 b2 or x2 a2 + y2 b2 = 1. This is the general equation of an ellipse with centre at O and foci at S and S'. The graph of an ellipse is as shown in Figure 5.7 below. O y + = 1 A'(–a, 0) S B'(0, –b) B(0, b) S A(a, 0) ' x x2 –– a2 y2 ––b2 Figure 5.7
Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 203 Notice that it is a closed curve with –a < x < a and –b < y < b. The curve intersects the x-axis at A(a, 0) and A'(–a, 0) and the y-axis at the points B(0, b) and B'(0, –b). Generally, a . b and AA' = 2a is called the major axis and BB' = 2b the minor axis. The points A and A' are the vertices of the ellipse. Since c = a2 – b2 , the foci are the points S(a2 – b2 , 0) and S'(–a2 – b2 , 0). If the centre of this ellipse is at the point (h, k), it can be shown that its equation is (x – h) 2 a2 + (y – k) 2 b2 = 1 The vertices are at A(h + a, k) and A'(h – a, k) and foci are at S(h + c, k) and S'(h – c, k), where c = a2 – b2 , as shown in Figure 5.8 below. O A' S' (h, k) S A y x + = 1 (x – h)2 –––––– a2 (y – k)2 –––––– b2 B' B Figure 5.8 The standard form of the equation of an ellipse with centre at the origin O is x2 a2 + y2 b2 = 1, where a = semi-major axis and b = semi-minor axis. The standard form of the equation of an ellipse with centre at the point (h, k) is (x – h) 2 a2 + (y – k) 2 b2 = 1 Note: In the special case of when b = a, the ellipse becomes a circle of radius a. Equation of a circle with radius a and centre at the origin O is x2 + y2 = a2 . Equation of a circle with radius a and centre at the point (h, k) is (x – h) 2 + (y – k) 2 = a2 . Example 9 State the coordinates of the centre, vertices and foci of the following ellipses: (a) x2 16 + y2 9 = 1 (b) (x + 2)2 25 + (y – 1)2 16 = 1 Solution: (a) Comparing x2 16 + y2 9 = 1 with the standard equation x2 a2 + y2 b2 = 1, we have a = 4, b = 3. Let c be the distance of the foci from the centre. From a2 = b2 + c 2 , c 2 = 16 – 9 = 7 c = 7 Hence, the centre is at the origin O, vertices are at (4, 0) and (–4, 0). The foci are at (7 , 0) and (–7 , 0).
204 Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 (b) Comparing (x + 2)2 25 + (y – 1)2 16 = 1 with the equation (x – h) 2 a2 + (y – k) 2 b2 = 1, we have h = –2, k = 1, a = 5 and b = 4. c = 25 – 16 = 9 = 3. Hence, the centre is at the point (–2, 1), vertices are at (3, 1) and (–7, 1). The foci are at (1, 1) and (–5, 1). Example 10 For each of the following ellipses, state the length of the major and minor axes. Sketh the curves. (a) x2 + 2y2 = 4 (b) 4x2 + y2 = 8 Solution: (a) Given that x2 + 2y2 = 4 x2 4 + y2 2 = 1 i.e. x2 22 + y2 ( 2 ) 2 = 1 This is the equation of an ellipse with a = 2 and b = 2. Thus, the major axis is 2a = 4 and the minor axis is 2b = 2 2. The sketch of the curve is shown on the right. (b) Given that 4x2 + y2 = 8 x2 2 + y2 8 = 1 i.e. x2 ( 2 ) 2 + y2 (2 2 ) 2 = 1 This is the equation of an ellipse with a = 2 and b = 2 2 . (Notice that in this case, b . a) Thus, the major axis is 2b = 4 2 , the minor axis is 2a = 2 2 . The sketch of the curve is shown on the right. Example 11 The centre of an ellipse is at the point (3, 2). The major axis is of length 12 units and parallel to the y-axis. The minor axis is of length 8 units and parallel to the x-axis. Find the equation of the ellipse. State the coordinates of its vertices and foci. Solution: The standard equation of an ellipse centre O, with major axis parallel to the y-axis and minor axis parallel to the x-axis is x2 a2 + y2 b2 = 1. Rewrite in the form x 2 a2 + y 2 b2 = 1 y x (–2, 0) O (2, 0) (0, – 2) (0, 2) y x O (0, –2 2) (0, 2 2) (– 2, 0) ( 2, 0)
Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 205 Hence, for the ellipse with centre at the point (3, 2) and major axis parallel to the y-axis with a = 4 and b = 6, its equation is (x – 3)2 42 + (y – 2)2 62 = 1. i.e. (x – 3)2 16 + (y – 2)2 36 = 1 The coordinates of the vertices are (3, 2 ± 6), i.e. (3, 8) and (3, –4). Distance of foci from centre is c = 36 – 16 = 20 = 2 5 . The coordinates of the foci are (3, 2 ± 2 5 ), i.e. (3, 2 + 2 5 ) and (3, 2 – 2 5 ). Parametric equations The ellipse x2 a2 + y2 b2 = 1 can also be defined by parametric equations. Let x = a cos q and y = b sin q, where q is a parameter, 0 < q < 2π. Then, x2 a2 + y2 b2 = a2 cos2 q a2 + b2 sin2 q b2 = cos2 q + sin2 q = 1 Hence, the equation of the locus of the point P can also be written as x = a cos q, y = b sin q Any point on the ellipse has parametric coordinates (a cos q, b sin q). Example 12 Find the parametric coordinates of any point on each of the following ellipses with cartesian equations (a) 4x2 + 9(y + 1)2 = 36 (b) (x + 2)2 + 4(y – 1)2 = 4 Solution: (a) For 4x2 + 9(y + 1)2 = 36, rewriting it in standard form x2 32 + (y + 1)2 22 = 1 The centre of the ellipse is at (0, –1), a = 3 and b = 2. The parametric equations are x = 3 cos q, y + 1 = 2 sin q. The parametric coordinates of any point are (3 cos q, –1 + 2 sin q). (b) For (x + 2)2 + 4(y – 1)2 = 4, rewriting in standard form (x + 2)2 22 + (y – 1)2 12 = 1 The centre of the ellipse is at (–2, 1), a = 2 and b = 1. The parametric equations are x + 2 = 2 cos q, y – 1 = sin q. The parametric coordinates of any point are (–2 + 2 cos q, 1 + sin q).
206 Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 Chords, tangents and normals to an ellipse Let x = a cos q, y = b sin q be the parametric equations of an ellipse x2 a2 + y2 b2 = 1. Let P and Q be two points on the ellipse with parametric coordinates (a cos q, b sin q) and (a cos f, b sin f) respectively. The gradient of the chord PQ = b(sin q – sin f) a(cos q – cos f) = b[2 cos 1 2 (q + f) sin 1 2 (q – f)] a[–2 sin 1 2 (q + f) sin 1 2 (q – f)] = – b a cos 1 2 (q + f) sin 1 2 (q + f) = – b a cot 1 2 (q + f) The equation of the chord PQ is y – b sin q = – b a cot 1 2 (q + f)(x – a cos q) y x B A' S' O S A B' P Q tangent chord normal Figure 5.9 Now, as the point Q moves towards P, the chord PQ gets shorter, until, when f = q, the point Q coincides with the point P. The chord PQ now becomes a tangent to the ellipse at P(a cos q, b sin q) with gradient – b a cot q. Hence, the equation of the tangent at P is y – b sin q = – b a cot q(x – a cos q) y – b sin q = – b cos q a sin q (x – a cos q) ay sin q – ab sin2 q = –bx cos q + ab cos2 q ay sin q + bx cos q = ab(sin2 q + cos2 q) ay sin q + bx cos q = ab
Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 207 Hence, x cos q a + y sin q b = 1 The gradient of the normal at P is a b tan q. The equation of the normal at P(a cos q, b sin q) is y – b sin q = a b tan q(x – a cos q) by cos q – b2 sin q cos q = ax sin q – a2 sin q cos q ax sin q – by cos q = (a2 – b2 ) sin q cos q Hence, ax cos q – by sin q = a2 – b2 Equation of the tangent at the point (a cos q, b sin q) is x cos q a + y sin q b = 1. Equation of the normal at the point (a cos q, b sin q) is ax cos q – by sin q = a2 – b2 . Example 13 Find the parameter q of the points with x-coordinate 2, and which lie on the ellipse 9x2 + 16y2 = 144. Hence, find the equation of the tangents to the ellipse at these points. Solution: For the ellipse 9x2 + 16y2 = 144, its standard form of equation is x2 16 + y2 9 = 1. Here a = 4 and b = 3. The parametric coordinates of the point are (4 cos q, 3 sin q). Hence, 4 cos q = 2 ⇒ cos q = 1 2 , q = ± π 3 . Equation of tangent at the point where q = π 3 is 1 4 x cos π 3 + 1 3 y sin π 3 = 1, i.e. 1 4 x · 1 2 + 1 3 y · 3 2 = 1 i.e. x 8 + 3 6 y = 1 3x + 43y = 24. Equation of tangent at the point where q = – π 3 is 1 4 x cos(– π 3 ) + 1 3 y sin(– π 3 ) = 1, i.e. 1 4 x · 1 2 – 1 3 y · 3 2 = 1 i.e. x 8 – 3 6 y = 1 3x – 43y = 24. Hence, equation of the tangents at the points where x = 2 are 3x + 43y = 24 and 3x – 43y = 24.
208 Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 Example 14 Find the equations of the tangents with gradient 2 to the ellipse with equation 2x2 + 3y2 = 6, and find their points of contact. Solution: The equation of the line with gradient 2 is y = 2x + c, where c is a constant to be found. If this line meets the ellipse 2x2 + 3y2 = 6, 2x2 + 3 (2x + c) 2 = 6 2x2 + 3(4x2 + 4cx + c 2 ) = 6 2x2 + 12x2 + 12cx + 3c 2 – 6 = 0 14x2 + 12 cx + 3(c 2 – 2) = 0 ………… For a point of contact, the equation (1) has equal roots or b2 – 4ac = 0. Thus (12c) 2 – 4 (14)(3)(c 2 – 2) = 0 6c 2 – 7c 2 + 14 = 0 c 2 = 14 c = ± 14 Hence, the tangents to the ellipse are y = 2x ± 14 . The points of contact are given by substituting c = ± 14 into : When c = 14 , 14x2 + 12 14 x + 3(14 – 2) = 0 14x2 + 12 14 x + 36 = 0 ( 14 x + 6)2 = 0 x = – 6 14 = – 3 7 14 y = – 6 7 14 + 14 = 1 7 14 When c = – 14 , 14x2 – 12 14 x + 36 = 0 ( 14 x – 6)2 = 0 x = 6 14 = 3 7 14 y = 6 7 14 – 14 = – 1 7 14 Hence, for the tangent y = 2x + 14 , the point of contact is (– 3 7 14 , 1 7 14 ). For the tangent y = 2x – 14 , the point of contact is ( 3 7 14 , – 1 7 14 ). Exercise 5.3 1. Sketch the curves represented by each of the following equations. State the length of the major and minor axes in each case. (a) 4x2 + 9y2 = 16 (b) 8x2 + 4y2 = 9 (c) 4x2 25 + 4y2 9 = 1 (d) (x – 2)2 4 + y2 8 = 1 (e) (x – 1)2 16 + (y – 3)2 9 = 1 (f) 4x2 + 9y2 – 12x + 36y + 9 = 0
Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 209 2. Write down the equations of the following ellipses: (a) Centre O, major axis horizontal and of length 8, minor axis of length 6. (b) Centre (0, 1), major axis vertical and of length 6, minor axis of length 4. (c) Centre (2, 1), major axis horizontal and of length 9, minor axis of length 62 . (d) Focus (1, 0), major axis horizontal and of length 4, minor axis of length 23 . (e) Focus (2, 1), major axis horizontal and of length 12, minor axis of length 45 . 3. Obtain the cartesian equations of the loci of the following points. Sketch the loci. (a) (cos q, 2 sin q) (b) (4 cos q, 3 sin q) (c) (5 – 4 cos q, 2 + 3 sin q) (d) (cos q + 1, 2 sin q – 1) 4. Obtain the parametric equations for the following ellipses: (a) (x – 2)2 9 + (y – 3)2 4 = 1 (b) 3(x + 2)2 + 2(y – 1)2 = 6 (c) (x – 2)2 + 4(y + 1)2 = 4 (d) 3(x – 1)2 + 5(y + 1)2 = 30 5. Find the equation of the tangents to the following ellipses at the specified points: (a) x = 3 cos q, y = 2 sin q, at the point q = π 3 ; (b) x = 5 cos q, y = 3 sin q, at the point q = 3π 4 ; (c) x = 4 cos q, y = 3 sin q, at the point q = – 2π 3 . 6. Find the equation of the tangents and normals to the following ellipses at the specified points: (a) 2x2 + 3y2 = 30 at the point (3, 2); (b) x2 + 4y2 = 8 at the point (2, –1); (c) 9x2 + 4y2 = 40 at the point (–2, –1). 7. Show that the following lines are tangents to the given ellipse, and find their points of contact. (a) x + 2y = 4; x2 + 4y2 = 8 (b) 2x + y = 8; x2 12 + y2 16 = 1 (c) 5x + 3y = 28; 5x2 + y2 = 56 (d) 3x + 7y = 13; 3x2 + 14y2 = 26 8. Prove that if the line lx + my + n = 0 touches the ellipse b2 x2 + a2 y2 = a2 b2 , then a2 l 2 + b2 m2 = n2 . 9. Find the gradients of the tangents drawn from the point (4, 6) to the ellipse x2 + 12y2 = 48. Hence, find the equations of the tangents and their points of contact with the ellipse. 10. If m is the gradient of the tangent from the point (3, 2) to the ellipse 9x2 + 16y2 = 144, find a quadratic equation in m. By noting whether the roots of this equation are real or imaginary, determine if the point (3, 2) lies within the ellipse. 11. Find the equation of the tangents to the ellipse 4x2 + 9y2 = 36, which are equally inclined to the x-axis and the y-axis. 12. Show that the line y = x – 5 is a tangent to the ellipse 9x2 + 16y2 = 144. Find the equations of the tangents which are perpendicular to y = x – 5. 13. Find the acute angle between the two tangents which can be drawn from the point (1, 1) to the ellipse 4x2 + 9y2 = 1. 14. S is the focus, on the positive x-axis, of the ellipse x2 25 + y2 9 = 1 and P(5 cos q, 3 sin q) is a variable point on the ellipse. If SP is produced to Q so that PQ = 2PS, find the equation of the locus of Q as P moves on the ellipse. 15. Find the equation of the tangents to the ellipse 3x2 + 4y2 = 12 which are parallel to the chord 2x + 3y = 1.
210 Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 The hyperbola A hyperbola is the locus of a point the difference of whose distances from two fixed points, called the foci, is constant. The mid-point of the segment connecting the foci is called the centre. Let S and S' be the two fixed points on the x-axis with coordinates (c, 0) and (–c, 0) respectively. Hence, the origin O is the centre of the hyperbola. If P(x, y) is any point on the hyperbola, then the difference of the distances is given by PS' – PS = [x – (–c)]2 + (y – 0)2 – (x – c) 2 + (y – 0)2 = constant Let us suppose that this constant is 2a. Then, [x – (–c)]2 + (y – 0)2 – (x – c) 2 + (y – 0)2 = 2a (x + c) 2 + y2 – (x – c) 2 + y2 = 2a. Let A, A' be two points on the x-axis with coordinates (a, 0) and (–a, 0) respectively, where a , c. We see that AS' – AS = AS' – A'S' = AA' = 2a. Hence, A(a, 0) and A'(–a, 0) are points on the hyperbola. y x P(x, y) S'(–c, 0) A' O A S(c, 0) Figure 5.10 From above, we have (x + c) 2 + y2 – (x – c) 2 + y2 = 2a (x + c) 2 + y2 = 2a + (x – c) 2 + y2 Squaring both sides, we have (x + c) 2 + y2 = 4a2 + (x – c) 2 + y2 + 4a (x – c) 2 + y2 x2 + 2cx + c 2 + y2 = 4a2 + x2 – 2cx + c 2 + y2 + 4a(x – c) 2 + y2 4a(x – c) 2 + y2 = 4cx – 4a2 a(x – c) 2 + y2 = cx – a2 Squaring both sides, we have a2 [(x – c) 2 + y2 ] = (cx – a2 ) 2 a2 (x2 – 2cx + c 2 + y2 ) = c 2 x2 – 2a2 cx + a4 a2 x2 – 2a2 cx + a2 c 2 + a2 y2 = c 2 x2 – 2a2 cx + a4 x2 (a2 – c 2 ) + a2 y2 = a2 (a2 – c 2 ) or x2 (c 2 – a2 ) – a2 y2 = a2 (c 2 – a2 ), since a , c. Let b be the positive number such that b2 = c 2 – a2 .
Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 211 Hence, b2 x2 – a2 y2 = a2 b2 or x2 a2 – y2 b2 = 1. By rearranging the terms to make y2 the subject, y2 = b2 a2 (x2 – a2 ) Since y2 > 0, x2 – a2 > 0 i.e. x < –a or x > a Hence, the curve does not exist for –a , x , a. Notice that for large values of x, either positive or negative. i.e. as x → ±∞, y 2 → b2 a2 x2 i.e. y → ± b a x Hence, y = ± b a x are the asymptotes of the curve x2 a2 – y2 b2 = 1. The graph of the hyperbola is as shown in Figure 5.11 below. Afi (–a, 0) A (a, 0) O y x b a y = – – x b a y = – x – – – = 1 x 2 y 2 a2 b2 Figure 5.11 Notice that it has two branches, with two asymptotes y = b a x and y = – b a x. The graph is symmetrical about both the x and y axes. The hyperbola has two foci at S(c, 0) and S'(–c, 0), where c = a2 + b2 . A(a, 0) and A'(–a, 0) are the vertices, AA' is the major axis and the mid-point of AA' is the centre of the hyperbola. If the centre of the hyperbola is at the point (h, k), it can be shown that its equation is (x – h) 2 a2 – (y – k) 2 b2 = 1 Its vertices are now at (h + a, k) and (h – a, k).
212 Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 Let X = x – h and Y = y – k, where the X-Y axes pass through (h, k) as the origin. The equation of the hyperbola is now X2 a2 – Y2 b2 = 1 The equations of the asymptotes with respect to the X-Y axes are Y = ± b a X. Hence, y – k = ± b a (x – h) or y = k ± b a (x – h) y x O y = k – — (x – h) b a (h, k) A(h + a, k) y = k + — (x – h) b a A'(h – a, k) (y – k)2 (x – h) 2 b2 a ——— – ——— = 1 2 Figure 5.12 The standard form of the equation of a hyperbola with centre at the origin O is x2 a2 – y2 b2 = 1, with asymptotes y = ± b a x. The standard form of the equation of a hyperbola with centre at the point (h, k) is (x – h) 2 a2 – (y – k) 2 b2 = 1, with asymptotes y = k ± b a (x – h). Example 15 The equation of a hyperbola is x2 16 – y2 9 = 1. Find (a) the coordinates of the foci of the hyperbola. (b) the equations of the asymptotes. Solution: (a) Comparing x2 16 – y2 9 = 1 with the standard equation x2 a2 – y2 b2 = 1, we have a = 4 and b = 3. Using c 2 = a2 + b2 , we have c 2 = 16 + 9 = 25 c = 5 The coordinates of the foci are (5, 0) and (–5, 0). (b) The equation of the asymptotes are y = ± b a x, i.e. y = ± 3 4 x.
Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 213 Example 16 Sketch the graph of each of the following curves. (a) x2 9 – y2 4 = 1 (b) x2 – y2 = 5 Solution: (a) x2 9 – y2 4 = 1, a = 3, b = 2. When y = 0, x2 = 9 x = ±3 Asymptotes are y = ± 2 3 x. The graph of x2 9 – y2 4 = 1 is as shown on the right. (b) x2 – y2 = 5 When y = 0, x2 = 5 x = ± 5 This is a rectangular hyperbola with asymptotes at y = ±x. The graph of x2 – y2 = 5 is as shown on the right. Example 17 The equation of a hyperbola is x2 – 2y2 – 6x – 4y – 11 = 0. (a) Obtain the standard form for the equation of the hyperbola. (b) Find the vertices, foci and the equations of the asymptotes of the hyperbola. Solution: (a) For the hyperbola x2 – 2y2 – 6x – 4y – 11 = 0. rewriting (x2 – 6x) – 2(y2 + 2y) – 11 = 0 (x – 3)2 – 2(y + 1)2 = 11 + 9 – 2 = 18 (x – 3)2 18 – (y + 1)2 9 = 1 or (x – 3)2 (3 2)2 – (y + 1)2 32 = 1. (b) We have a = 3 2 and b = 3. Using c 2 = a2 + b2 , c 2 = 18 + 9 = 27 c = 3 3 The centre of the hyperbola is at (3, –1). The vertices are at (3 ± 3 2 , –1). The foci are at (3 ± 3 3 , –1). The asymptotes are y + 1 = ± 1 2 (x – 3) i.e. 2 (y + 1) = x – 3 and 2 (y + 1) = 3 – x. (–3, 0) (3, 0) 0 y x – – – = 1 x 2 y 2 9 4 y = – – x 2 3 y = – x 2 3 0 y x y = x y = – x x 2 – y 2 = 5 (–fi5, 0) (fi5, 0)
214 Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 Parametric equations The hyperbola x2 a2 – y2 b2 = 1 can also be defined by parametric equations. Let x = a sec q and y = b tan q, where q is a parameter. Then, x2 a2 – y2 b2 = a2 sec2 q a2 – b2 tan2 q b2 = sec2 q – tan2 q = 1 Hence, the equation of the locus of the point P can also be written as x = a sec q, y = b tan q. Any point on the hyperbola has parametric coordinates (a sec q, b tan q). Chords, tangents and normals to a hyperbola Let x = a sec q, y = b tan q be the parametric equation of a hyperbola x2 a2 – y2 b2 = 1. Let P and Q be two points on the ellipse with parametric coordinates (a sec q, b tan q) and (a sec f, b tan f) respectively. The equation of the chord PQ is y – b tan q = b(tan q – tan f) a(sec q – sec f) (x – a sec q) y b (sec q – sec f) – tan q(sec q – sec f) = x a (tan q – tan f) – sec q(tan q – tan f) x a (tan q – tan f) – y b (sec q – sec f) = tan q sec f – sec q tan f Multiplying both sides of the equation by cos q cos f, we have x a (sin q cos f – cos q sin f) – y b (cos f – cos q) = sin q – sin f x a sin(q – f) – y b (cos f – cos q) = sin q – sin f x a 32 sin 1 2 (q – f) cos 1 2 (q – f)4 – y b 32 sin 1 2 (q + f) sin 1 2 (q – f)4 = 2 cos 1 2 (q + f) sin 1 2 (q – f) x a cos 1 2 (q – f) – y b sin 1 2 (q + f) = cos 1 2 (q + f) Now, as the point Q moves towards P, the chord PQ gets shorter, until, when f = q, the point Q coincides with the point P. The chord PQ now becomes a tangent to the hyperbola at P(a sec q, b tan q). Hence, the equation of the tangent at the point P(a sec q, b tan q) is x a – y b sin q = cos q or x sec q a – y tan q b = 1 Gradient of the tangent at P(a sec q, b tan q) is b a sin q . Hence, gradient of the normal at P(a sec q, b tan q) is – a sin q b .
Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 215 Equation of the normal at the point P(a sec q, b tan q) is y – b tan q = – a sin q b (x – a sec q) by – b2 tan q = –ax sin q + a2 tan q ax sin q + by = (a2 + b2 ) tan q i.e. ax sec q + by tan q = a2 + b2 Example 18 Show that the point 1 3 5 2 , 22 lies on the hyperbola x2 9 – y2 16 = 1. By finding its parametric coordinates, find the equation of the tangent to the hyperbola x2 9 – y2 16 = 1 at the point 1 3 5 2 , 22. Solution: Equation of the hyperbola is x2 9 – y2 16 = 1 i.e. y2 = 161 x2 9 – 12 When x = 3 5 2 , y2 = 161 5 4 – 12 = 4 ⇒ y = 2 Therefore, the point 1 3 5 2 , 22 lies on the hyperbola x2 9 – y2 16 = 1. If its parametric coordinates are (a sec q, b tan q), where a = 3 and b = 4. we have 3 sec q = 3 5 2 and 4 tan q = 2. Hence, sec q = 5 2 and tan q = 1 2 . Equation of the tangent at the point (a sec q, b tan q) is x sec q a – y tan q b = 1. Hence, equation of the tangent at the point 1 3 5 2 , 22 is x 3 · 5 2 – y 4 · 1 2 = 1 i.e. 4 5x – 3y = 24 Equation of the tangent at the point P(a sec q, b tan q) is x sec q a – y tan q b = 1. Equation of the normal at the point P(a sec q, b tan q) is ax sec q + by tan q = a2 + b2 .
216 Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 The rectangular hyperbola If a = b, the equation of the hyperbola x2 a2 – y2 b2 = 1 reduces to x2 – y2 = a2 . The equations of the asymptotes are y = ±x, inclined at ± π 4 to the axes, and are perpendicular to one another. This type of hyperbola is called the rectangular hyperbola. Let P(x, y) be a point on the rectangular hyperbola. Let (X, Y) be the coordinates of P with the asymptotes Ox and Oy as the coordinate axes (see Figure 5.13). Then, X = OQ = OM – QM = OM – LN = ON cos 45° – PN cos 45° i.e. X = x 2 – y 2 . Similarly, Y = PQ = PL + QL = x 2 + y 2 XY = 1 x 2 – y 2 21 x 2 + y 2 2 = x2 – y2 2 Since x2 – y2 = a2 , XY = 1 2 a2 . O y Y X x A A' N P L Q M Figure 5.13 Hence, if we rotate the rectangular hyperbola through an angle π 4 in the positive direction (anti-clockwise), then using the asymptotes as the new x and y-axes, the equation of the rectangular hyperbola is now xy = 1 2 a2 or xy = c 2 . The graphs of x2 – y2 = a2 and xy = c 2 are as shown in Figure 5.13 and Figure 5.14. O y x xy = c 2 Figure 5.14
Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 217 O (2, 2) (–2, –2) y x xy = c 2 –4 O y x x(y – 1) = 4 y = 1 A A' Example 19 Sketch the graph of each of the following curves: (a) xy = 4 (b) x(y – 1) = 4 State the coordinates of their vertices. Solution: (a) For xy = 4, c 2 = 4 = 1 2 a2 a = 2 2 The graph of xy = 4 is as shown on the right. The vertices are at (2, 2) and (–2, –2). (b) x(y – 1) = 4 Let X = x and Y = y – 1 Then XY = 4 is a rectangular hyperbola with asymptotes at X = 0, i.e. x = 0 and Y = 0, i.e. y = 1 When y = 0, x = –4 The graph of x(y – 1) = 4 is as shown on the right. The vertices are at A(2, 3) and A´(–2, –1). Parametric equations The rectangular hyperbola xy = c 2 can also be defined by parametric equations. Let x = ct and y = c t , where t R, t ≠ 0 is the parameter. Then, xy = ct · c t = c 2 Hence, the equation of the locus of the point P can be written as x = ct, y = c t , t R, t ≠ 0 Any point lying on the rectangular hyperbola can be represented by the parametric coordinates 1ct, c t 2.
218 Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 Example 20 Find the locus of the following points. (a) P12t, 2 t 2 (b) Q1 3 2 t, 3 2t 2 Solution: (a) Let x = 2t and y = 2 t Multiplying, xy = 2t · 2 t = 4 i.e. the locus of P is the rectangular hyperbola xy = 4 (b) Let x = 3 2 t and y = 3 2t Multiplying, xy = 3 2 t · 3 2t = 9 4 i.e. the locus of Q is the rectangular hyperbola 4xy = 9. Example 21 For each of the following curves, obtain its parametric equations. (a) 25xy = 64 (b) (x – 1)(y – 2) = 4 Solution: (a) 25xy = 64 xy = 64 25 = 1 8 5 2 2 ∴ The parametric equations are x = 8 5 t, y = 8 5t . (b) (x – 1)(y – 2) = 4 = (2)2 Let x – 1 = 2t and y – 2 = 2 t x = 1 + 2t, y = 2 + 2 t ∴ The parametric equations are x = 1 + 2t, y = 2 + 2 t . Chords, tangents and normals to a rectangular hyperbola Let P1cp, c p 2 and Q1cq, c q 2 be two points on the rectangular hyperbola xy = c 2 . c q – c p Then, gradient of PQ = cq – cp p – q pq = q – p = – 1 pq
Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 219 Hence, gradient of the chord joining the points P1cp, c p 2 and Q1cq, c q 2 is – 1 pq . Equation of the chord PQ is y – c p = – 1 pq (x – cp) pqy – cq = –x + cp i.e. pqy + x = c(p + q) Now, as Q moves towards P, the chord PQ gets shorter until when q = p, the point Q coincides with point P. The chord PQ becomes the tangent at P. Hence, gradient of tangent at P = – 1 p2 . Equation of tangent at P is p2 y + x = 2cp Gradient of normal at P is p2 . Example 22 P1cp, c p 2 and Q1cq, c q 2 are two points on the curve xy = c 2 . Find the equation of PQ. The chord PQ subtends a right angle at the point R on the curve. Prove that the normal at R is parallel to PQ. Solution: Equation of the chord PQ is pqy + x = c(p + q) Gradient of PQ = – 1 pq Let R be 1cr, c r 2 Then, gradient of PR = – 1 pr and gradient of QR = – 1 qr ∠PRQ = 90°, i.e. PR is perpendicular to QR. Thus 1– 1 pr 21– 1 qr 2 = –1 1 pqr 2 = –1 – 1 pq = r 2 ………… Gradient of tangent at R = – 1 r 2 . Gradient of normal at R = r 2 Hence, from , gradient of PQ = gradient of normal at R. i.e. PQ is parallel to normal at R. O y x P (cp, – ) c p Q (cq, – ) c q tangent at P Figure 5.15 P (cp, – ) c p R (cr, – ) c r Q (cq, – ) c q normal at R O y x
220 Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 Example 23 Show that the equation of the normal at the point P1ct, c t 2 on the hyperbola xy = c 2 is ty – t 3 x = c(1 – t 4 ). Hence, show that there exists four normals that can be drawn from the point (–5, –5) to the hyperbola xy = c 2 , and find the four points of intersection of the normals with the curve. Solution: Gradient of tangent at P1ct, c t 2 is – 1 t 2 . Gradient of normal at P is t 2 . Equation of normal at P1ct, c t 2 is y – c t = t2 (x – ct) ty – t 3 x = c(1 – t 4 ) ………… For the hyperbola xy = 4, c = 2. If the normal passes through the point (–5, –5), substitute x = –5, y = –5 and c = 2 into : –5t + 5t 3 = 2(1 – t 4 ) 2t 4 + 5t 3 – 5t – 2 = 0 ………… This is a quartic equation in terms of t. Hence, there are four values of t satisfying the equation . This shows that there are four points on the hyperbola from where the normals pass through the point (–5, –5). From , 2t 4 + 5t 3 – 5t – 2 = 0 (t – 1)(t + 1)(t + 2)(2t + 1) = 0 t = 1, –1, –2 or – 1 2 With these values of t and using c = 2, the four points on the hyperbola are (–2, –2), (2, 2), (– 4, –1), and (–1, – 4). y x xy = 4 O (2, 2) (–2, –2) (–1, –4) (–5, –5) (–4, –1)
Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 221 Exercise 5.4 1. Sketch the graph of each of the following curves, showing clearly the asymptotes. (a) x2 16 – y2 9 = 1 (b) x2 – 2y2 = 10 (c) xy = 9 (d) 2xy = 15 (e) x = 2 y – 1 (f) y = 2 x – 2 (g) xy = –16 (h) (x – 1)(y – 2) = 1 2. For each of the following hyperbolas, find (i) the coordinates of the foci, (ii) equations of the directrices and the asymptotes. (a) x2 – 4y2 = 4 (b) x2 – 3y2 = 12 (c) x2 64 – y2 16 = 1 (d) x2 36 – y2 9 = 1 (e) 4(x – 1)2 – 9y2 = 36 (f) (x – 1)2 45 – (y – 2)2 25 = 1 3. Find the cartesian equations of the curves having the following parametric equations. (a) x = 3t, y = 3 t (b) x = 5 2 t, y = 5 2t (c) x = 2t, y = – 2 t (d) x = – 2 3 t, y = 2 3t (e) x = 1 + t, y = 1 t (f) x = 4t, y = 1 – 4 t 4. Write down the parametric equations of the curves having the following cartesian equations. (a) xy = 16 (b) 9xy = 25 (c) 16xy = –1 (d) y = 64 x (e) x – 1 = 4 y (f) (x – 2)y = 1 5. If the gradient of the tangent at the point 1ct, c t 2 on the hyperbola xy = c 2 is – 1 t 2 , find the equations of the tangents and normals at each of the given points on the given hyperbola below. (a) 13t, 3 t 2; xy = 9 (b) 13, 4 3 2; xy = 4 (c) 1 3 2 t, 3 2t 2; 4xy = 9 (d) 1 9 2 , 3 4 2; 8xy = 27 6. Find the equation of the tangent to the hyperbola xy = 4 satisfying the following: (a) it has a gradient – 3 4 (b) it passes through the point (2, 0) 7. The normal at the point P16, 3 2 2 which lies on the hyperbola x = 3t and y = 3 t meets the curve again at Q. Find the coordinates of Q and the length of PQ. 8. The tangent and normal at the point P(3, 4) which lies on the curve xy = 12, meet the x and y-axes respectively at Q, R and Q, R. Find the length of QR and QR. 9. P and Q are two variable points lying on the hyperbola x = 3t, y = 3 t . The tangents at P and Q meet at T. If PQ passes through the point (6, 2), find the equation of the locus of T as PQ varies. 10. The line 2y = x + 7 intersects the curve x = 2t, y = 2 t at A and B. Find the respective values of t corresponding to A and B. Hence, find the coordinates of the point of intersection of the tangents at A and B. 11. Prove that for all values of k, the line k2 x + y = 8k is a tangent to the hyperbola xy = 16. Find the coordinates of the point of contact.
222 Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 12. By finding the gradients of the tangents from the point (2, –3) to the hyperbola 4xy = 25, find the acute angle between the two tangents. 13. The normal to the hyperbola xy = c 2 at the point Q1cq, c q 2 intersects the straight line y = x at R. If O is the origin, show that OQ = QR. If the tangent to the hyperbola at Q intersects OR at P, prove that OP · OR = 4c 2 . 14. If the line y = 2x + 1 is a tangent to the hyperbola x2 – py2 = 1, find the value of p. 15. The line y = mx + c touches the hyperbola x2 a2 – y 2 b2 = 1. Find the possible values of m. Hence find the equations of the tangents from the origin to the hyperbola. Summary 1. The standard form of the equation of a circle with centre (a, b) and radius r is (x – a) 2 + (y – b) 2 = r 2 . 2. The standard form of the equation of an ellipse with centre (h, k) and foci (h – c, k), (h + c, k), where c 2 = a2 – b2 , is (x – h) 2 a2 + (y – k) 2 b2 = 1. The major axis is 2a and the minor axis is 2b. 3. The parametric equations of the ellipse x2 a2 + y 2 b2 = 1 are x = a cos q and y = b sin q. 4. The standard form of the equation of a parabola with vertex (h, k), focus (a + h, k) and directrix x = –a + h is (y – k) 2 = 4a(x – h) 5. The parametric equations of the parabola y2 = 4ax are x = at2 and y = 2at. 6. The standard form of the equation of a hyperbola with centre (h, k) and foci (h – c, k), (h + c, k), where c 2 = a2 + b2 , is (x – h) 2 a2 – (y – k) 2 b2 = 1. The asymptotes to the hyperbola are y = k ± b a (x – h). 7. The parametric equations of the hyperbola x2 a2 – y 2 b2 = 1 are x = a sec q, y = b tan q. 8. The equation of a rectangular hyperbola with y = 0 (x-axis) and x = 0 (y-axis) as asymptotes is xy = c 2 . 9. The parametric equations of the rectangular hyperbola xy = c 2 are x = ct and y = c t .
Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 223 STPM PRACTICE 5 1. The general equation of a conic is 12x2 + 16y2 + 48x – 96y = 0. (a) Obtain the standard form for the equation and identify the conic. (b) Find the coordinates of the centre and foci of the conic. (c) Sketch the conic, and indicate its centre and foci on the graph. 2. The parametric equations of a conic are x = a sin q + 9 and y = b cos q – 12, where a and b are positive constants and 0 < q < 2π. (a) Find the standard form for the equation of the conic, and identify the type of conic. (b) If a = b = 15, determine and sketch the conic. 3. P(ap2 , 2ap) is any point on the parabola y2 = 4ax. S(a, 0) is the focus of the parabola and a straight line through S and perpendicular to SP meets the tangent to the parabola at P at the point L. Prove that L lies on the directrix x = –a of the parabola. Show that the area of the triangle PSL is a2 (1 + p2 ) 2 2p . 4. Prove that for all values of m, the line y = mx + a m is a tangent to the parabola y2 = 4ax. Hence, find a quadratic equation whose roots are the gradients of the tangents from the point (x, y) to the parabola. Prove that, if the tangents from a point P to the parabola are inclined at an angle to one another, P must lie on the curve with equation (x + a) 2 tan2 = y 2 – 4ax. 5. A parabola is represented by the parametric equations x = 4t 2 , y = 8t. P and Q are two points, with nonzero parametric p and q respectively. Point A is the focus of the parabola. (a) Find the equation of the chord PQ in terms of p and q. Hence, deduce the equation of the tangent to the parabola at point P. (b) Find the coordinates of M, where the tangent to the parabola at point P cuts the x-axis. (c) State the coordinates of A and show that AM = AP. (d) If the normal to the parabola at point P cuts the parabola curve again at point Q, show that p2 + pq + 2 = 0. 6. The straight line y = mx + c intersects the ellipse x2 + 4y2 = 16 at the points P and Q. Show that the coordinates of M, the mid-point of PQ, are x = – 4mc 4m2 + 1 , y = c 4m2 + 1 If the chord PQ passes through the point (2, 0), show that M lies on the ellipse x2 + 4y2 = 2x. Sketch both the ellipses on the same diagram. 7. Prove that the line y = mx + c is a tangent to the ellipse b2 x2 + a2 y2 = a2 b2 if a2 m2 = c 2 – b2 . Show that if the line y = mx + c passes through the point 1 5 4 , 52 and is a tangent to the ellipse 8x2 + 3y2 = 35, then c = 35 3 or 35 9 . Find the coordinates of the point of contact of the tangent from the point 1 5 4 , 52 to the curve 8x2 + 3y2 = 35. 8. The equation of an ellipse is 16x2 + 9y2 – 32x – 72y + 16 = 0. (a) Obtain the standard form for the equation of the ellipse. (b) Find the coordinates of the centre C, the focus F1 and the focus F2 of the ellipse. (c) Sketch the ellipse, and indicate the points C, F1 and F2 on the ellipse.
224 Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 9. Prove that for all values of f the point (a cos f, b sin f) lies on the ellipse x2 a2 + y 2 b2 = 1, and obtain the equation of the tangent to the ellipse at this point. If this tangent intersects the coordinate axes at Q and R, and M is the mid-point of QR, find the coordinates of M, and hence show that as f varies, the locus of M is the curve a2 x2 + b2 y2 = 4. 10. Prove that the tangent to the hyperbola xy = c 2 at the point P1cp, c p 2 is p2 y + x = 2cp. The perpendicular line from the origin meets this tangent at N and meets the hyperbola again at Q and R. Prove that (a) ∠QPR is a right angle, (b) N lies on the curve (x2 + y2 ) 2 = 4c 2 xy. 11. Find the equation of the tangent to the hyperbola xy = c 2 at the point P1cp, c p 2. The tangent at P meets the x-axis at L and the y-axis at M. O is the centre of the hyperbola and POQ is a diameter. The straight line MQ meets the x-axis at T. Prove that (a) area of the triangle MOL is 2c 2 , (b) area of the triangle QOT is c 2 3 . 12. Points P1cp, c p 2, Q1cq, c q 2 and R1cr, c r 2 lie on the rectangular hyperbola xy = c2 . (a) If chord PQ is perpendicular to chord QR, show that pq2 r = –1. (b) Show that the chord PQ has the equation pqy + x = c(p + q). Hence, by deduction, find the equation of the tangent to the hyperbola at Q. (c) The tangent at point Q cuts the x-axis at point M and cuts the y-axis at point N. If O is an origin, show that the area of the triangle OMN is 2c 2 . 13. The equation of a hyperbola is x2 16 – y 2 9 = 1. (a) Find the coordinates of the foci. (b) Determine the equations of the asymptotes. 14. A variable point P on the hyperbola x2 a2 – y 2 b2 = 1 is joined to the vertex A(a, 0). If Q is the mid-point of PA, show that the locus of Q is another hyperbola with centre at 1 a 2 , 02. 15. The point P(a sec t, b tan t) on the hyperbola x2 a2 – y 2 b2 = 1 is joined to the vertices A(a, 0) and B(–a, 0). The lines AP, BP meet the asymptote ay = bx at Q, R respectively. Prove that the x-coordinate of Q is 1a cos 1 2 t2 / 5cos 1 2 t – sin 1 2 t6, and that the length of QR is independent of the value of t. 16. Show that the equation of the tangent at the point P(a sec a, a tan a) on the hyperbola x2 – y2 = a2 is x – y sin a = a cos a. If the perpendicular from the origin O on the tangent meets the tangent at N and meets the hyperbola at Q, prove that ON. OQ = a2 . 17. The equation of a hyperbola is y2 – x2 – 6y = 0. (a) Find the standard form of the equation of the hyperbola. State the coordinates of the centre and vertices of the hyperbola. (b) Determine the equations of the asymptotes of the hyperbola, and show that the asymptotes are perpendicular to each other. (c) Sketch the hyperbola.
Mathematics Term 1 STPM Chapter 6 Vectors 6 CHAPTER 6 VECTORS Subtopic Learning Outcome 6.1 Vectors in two and three dimensions (a) Use unit vectors and position vectors. (b) Perform scalar multiplication, addition and subtraction of vectors. (c) Find the scalar product of two vectors, and determine the angle between two vectors. (d) Find the vector product of two vectors, and determine the area of a parallelogram and of a triangle. 6.2 Vector geometry (a) Find and use the vector and cartesian equations of lines. (b) Find and use the vector and cartesian equations of planes. (c) Calculate the angle between two lines, between a line and a plane, and between two planes. (d) Find the point of intersection of two lines, and of a line and a plane. (e) Find the line of intersection of two planes. angle – sudut course – haluan direction – arah line – garis magnitude – magnitud plane – satah position vector – vektor kedudukan resultant velocity – halaju paduan scalar – skalar scalar product – hasil darab skalar speed – laju unit vector – vektor unit vector – vektor vector product – hasil darab vektor Bilingual Keywords
226 Mathematics Term 1 STPM Chapter 6 Vectors 6 6.1 Vectors in Two and Three Dimensions Vectors To describe each of the physical quantities like velocity, acceleration, force, momentum and especially displacement, we must specify both its magnitude and the direction. All these physical quantities that have both magnitude and direction are known as vectors. Geometrically, we represent a vector as a directed line segment, that is a line segment to which a direction has been assigned. T S Figure 6.1 A directed line segment from the initial point S to the terminal point T is represented by →ST. The length of the line segment is its magnitude. In notation, a vector can be written in two ways: (a) Two capital letters with an arrow on top such as →ST and →AB. Such vectors are called displacement vectors. (b) A small letter either in boldprint or with tilde (~) beneath such as a or a ~. In both cases, the magnitude of the vector is written in modulus form |→ST|, |a| or |a ~|. Zero vector (Null vector) A vector with magnitude 0 and indeterminate direction. Opposite vectors Two vectors with equal magnitude but in opposite directions. B C A D Figure 6.2 | →AB| = | →CD| but →AB = – →CD Equal vectors Vectors which have the same magnitude and in the same directions. W V S R Q P Figure 6.3 →PQ = →RS = →VW because | →PQ| = | →RS | = | →VW| and →PQ // →RS // →VW
Mathematics Term 1 STPM Chapter 6 Vectors 6 227 Parallel vectors Vectors which have different magnitudes but can have the same or opposite directions. L M K J FG E H Figure 6.4 →EF , →GH, →JK and →LM are all parallel vectors. Unit vector A vector of magnitude 1 unit. Q 1 unit P Figure 6.5 →PQ is a unit vector since | →PQ| = 1 Unit vector is the direction of →PQ is defined as →PQ | →PQ| . Any non zero vectors, when divided by its own magnitude will produce a unit vector which is in the direction of the vector itself. Position vectors Position vectors are displacement vectors which begin from the origin. x B A R O P y Figure 6.6 →OA, →OB, →OP and →OR are all position vectors since their initial point are located at the origin (0, 0). Vectors can exist in many dimensions. Two dimensional vectors or coplanar vectors are described by ordered pairs of cartesian coordinates (x, y) and drawn on xy-plane. Besides, we have vectors described by ordered triples of cartesian coordinates (x, y, z) and drawn on xyz-plane known as three dimensional vectors or space vectors.
228 Mathematics Term 1 STPM Chapter 6 Vectors 6 In two dimension, vector a can be represented by column vectors, linear combination of i and j, a = 1 p q 2 or a = pi + qj, where i = 1 1 o 2 and j = 1 0 1 2 are base vectors. x A (p, q) i a j (1, 0) (0, 1) O q p y Figure 6.7 i and j are also known as the unit vectors in the direction of positive x and y respectively. so, i = (1, 0); |i| = 1 j = (0, 1); |j| = 1 The vector from O to A is referred as the position vector of A, written as OA = 1 p q 2 a column vector, or in the form of linear combination of unit vectors i and j, →OA = pi + qj. THe magnitude of position vector →OA is equivalent to the length of the line OA which is = | →OA| = p2 + q2 In three dimensions, vector a can be written in the form of column vector linear combination of i, j and k. a = 1 p q r 2 or a = pi + qj + rk, where i = 1 1 0 0 2, j = 1 0 1 0 2 and k = 1 0 0 1 2. i, j and k are the unit vectors in the direction of positive x, y and z axes respectively. y x j k i (0, 0, r) (0, q, 0) A (p, q, r) (p, 0, 0) (0, 0, 1) (1, 0, 0) (0, 1, 0) z so, i = (1, 0, 0); |i| = 1 j = (0, 1, 0); |j| = 1 k = (0, 0, 1); |k| = 1 Figure 6.8 Vectors INFO
Mathematics Term 1 STPM Chapter 6 Vectors 6 229 The vector from O to A is also referred as the position vector of A, written as →OA = 1 p q r 2 or in the form of →OA = pi + qj + rk. The magnitude of position vector →OA is also equivalent to the length of the line OA which is = | →OA| = p2 + q2 + r2 Scalars Scalar is a physical quantity which is represented by a single real value (magnitude) only. Scalar does not have a direction. Physical quantities such as distance, length, time, temperature, volume and density are examples of scalars which can be specified completely by magnitude only. Example 1 Plot the points: (a) P(2, 4, 6) (b) Q(6, –4, 8) (c) R(–2, 6, –4) Solution: (a) P(2, 4, 6) O y x z 2 4 6 –2 –4 P(2, 4, 6) 2 2 2 4 4 4 6 (b) Q(6, –4, 8) O y x z 2 4 6 8 10 –2 –4 2 4 6 2 Q(6, –4, 8)
230 Mathematics Term 1 STPM Chapter 6 Vectors 6 (c) R(–2, 6, –4) O y R(–2, 6, –4) x z 2 –4 4 –2 –2 –4 2 –2 2 4 6 8 4 Example 2 If the coordinates of P is (1, –2, 4), find (a) the position vector of P, (b) the magnitude of →OP. Solution: (a) →OP = i – 2j + 4k (b) | →OP| = 12 + (–2)2 + (4)2 | →OP| = 21 Exercise 6.1 1. Find a unit vector which is in the same direction as the given vector. (a) 4i – 3j (b) 5i + 12j (c) –2i – j 2. Find a vector of magnitude 13 which has the opposite direction of –12i + 5j. 3. If v = 2i + 3j – 6k, find (a) a unit vector having the same direction as v. (b) a vector having twice the magnitude and in the opposite direction of v. 4. Given that →OA = –i + 3j + 2k and →OB = 4i + j + 3k where O is the origin. Find the vector →AB and the magnitude of →AB. 5. Given that →OP = 6i + 2j – k and →OQ = 2i + 3j + 2k, find a unit vector in the direction of →PQ. 6. The point P has position vector 3i – j + 5k. Given that →PQ = 3i + 4j + 2k. Find the position vector of the point Q. 7. Find the length of the vector →OA if A is the point (a) (2, 4, –1) (b) (4, 0, –3) (c) (1, –2, 2)
Mathematics Term 1 STPM Chapter 6 Vectors 6 231 8. Find the magnitude of the vector v if (a) v = i – 2j + k (b) v = –3i + 2j + 6k (c) v = 7i – 11j – 4k 9. Given sets of points A, B and C. Determine whether the point C lies on the line AB. (a) A(3, 2, 4), B(–3, –7, –8), C(0, 1, 3) (b) A(–3, 1, 3), B(3, 1, 0), C(5, 1, –1) (c) A(–2, 9, 1), B(–5, 3, 4), C(–4, 5, 3) 10. Show that the three points whose position vectors are 2i + 3j – 4k, i – 2j + 3k and –7j + 10k are collinear. 11. Points R, S and T have coordinates (1, 3, 2), (3, 1, 4) and (4, 1, –5) respectively. Find (a) →RS and →ST in terms of i, j and k (b) 2 →RS – 1 2 →RT + 1 2 →ST in terms of i, j and k. Algebraic operations on vectors Scalar multiplication of a vector Multiplying a non zero vector by a scalar will produce another vector. The magnitude of the vector changes and the direction of the vector depends on the nature of the scalar. When vector a is multiplied by a scalar l, we will get la. a and la are two parallel vectors. la is a vector with magnitude l times the magnitude of a. (a) If l is positive, i.e. l . 0, then la and a are parallel in the same direction. ( = 1) fi a ( = 2) fi 2a 1 2 1 2 fi = –ff fi –a Figure 6.9(a) (b) If l is negative, i.e. l , 0, then la and a are parallel in the opposite direction. ( λ = 1) a (λ = –2) –2a (λ = –1) –a 1 2 1 2 fiff = – –ff – –a Figure 6.9(b) (c) If l = 0, then la is a zero vector. (d) If two vectors a and b are parallel, then vector a can be written as a scalar multiple of vector b. a = lb where l is a scalar (i) If l is a positive scalar, l . 0, a b Figure 6.10(a)
232 Mathematics Term 1 STPM Chapter 6 Vectors 6 (ii) If l is a negative scalar, l , 0. a b Figure 6.10(b) When we multiply a column vector by a scalar, we have to multiply each component of the vector by the scalar. For example, If a = 1 x y 2 and b = 1 x y z 2 then 3a = 31 x y 2 = 1 3x 3y2 –5b = –51 x y z 2 = 1 –5x –5y –5z 2 Addition of vectors Two or more vectors can be added geometrically or algebraically to produce a single vector known as the resultant vector to represent all of the vectors. Geometrically, addition of vectors can be carried out by using the triangle or parallelogram rule. (a) Triangle rule of addition If two vectors p and q and positioned so that the initial point of q is at the terminal point of p, then the sum p + q is the vector from the initial point of p to the terminal point of q. p + = B A q p q r D C Figure 6.11(a) →AB + →CD = p + q = r (b) Parallelogram rule of addition If two vectors p and q have the same initial point, then the sum p + q is represented by the diagonal of the parallelogram with p and q as adjacent sides. q p p + q p q B C A D Figure 6.11(b) →AB + →BC = →AC = →AD + →DC Note that p + q = q + p. This is called the commutative rule for the addition of vectors.
Mathematics Term 1 STPM Chapter 6 Vectors 6 233 Algebraically, to add two vectors, just have to add each of the corresponding components of the vectors together. If →OA = 1 x1 y1 2 and →OB = 1 x2 y2 2, Then, →OA + →OB = 1 x1 y1 2 + 1 x2 y2 2 = 1 x1 + x2 y1 + y2 2 If →OP = 1 x1 y1 z1 2 and →OQ = 1 x2 y2 z2 2 Then, →OP + →OQ = 1 x1 y1 z1 2 + 1 x2 y2 z2 2 →OP + →OQ = 1 x1 + x2 y1 + y2 z1 + z2 2 Example 3 Find (a) 3 1 1 –2 6 2 + 21 0 4 –22, (b) 5 1 1 0 –22 + 31 –1 4 6 2 + 21 –4 –5 –62. Solution: (a) 3 1 1 –2 6 2 + 21 0 4 –22 = 1 3 –6 18 2 + 1 0 8 –42 = 1 3 + 0 –6 + 8 18 + (–4)2 = 1 3 2 14 2 (b) 5 1 1 0 –22 + 31 –1 4 6 2 + 21 –4 –5 –62 = 1 5 0 –102 + 1 –3 12 18 2 + 1 –8 –10 –122 = 1 5 – 3 – 8 0 + 12 + (–10) –10 + 18 – 12 2 = 1 –6 2 –42
234 Mathematics Term 1 STPM Chapter 6 Vectors 6 Example 4 Show that the vectors a = i + j + 3k, b = 2i + j – 5k and c = –3i – 2j + 2k form a triangle. Solution: Three vectors will form a triangle if the sum of all the vectors is zero. a + b + c = (i + j + 3k) + (2i + j – 5k) + (–3i – 2j + 2k) = (i + 2i – 3i) + (j + j – 2j) + (3k – 5k + 2k) a + b + c = 0 So, the three vector form a triangle. Subtraction of two vectors Subtracting a vector is the same as adding its inverse vector. p – q = p + (–q) p – q p p A B D C q –q Figure 6.12 The different p – q is the vector from the terminal point of q to the terminal point of p. To subtract algebraically, we have to subtract the corresponding component of the second vector. If →OA = 1 x1 y1 2 and →OB = 1 x2 y2 2. Then, →OA – →OB = 1 x1 y1 2 – 1 x2 y2 2 = 1 x1 – x2 y1 – y2 2 If →OP = 1 x1 y1 z1 2 and →OQ = 1 x2 y2 z2 2 Then, →OP – →OQ = 1 x1 y1 z1 2 – 1 x2 y2 z2 2 →OP – →OQ = 1 x1 – x2 y1 – y2 z1 – z2 2
Mathematics Term 1 STPM Chapter 6 Vectors 6 235 Example 5 Find (a) 2 1 1 –1 3 2 – 31 2 3 –42, (b) –21 5 3 4 2 – 41 –1 5 2 2 – 31 3 –1 0 2. Solution: (a) 2 1 1 –1 3 2 – 31 2 3 –42 = 1 2 –2 6 2 – 1 6 9 –122 = 1 2 – 6 –2 – 9 6 – (–12)2 = 1 –4 –11 18 2 (b) –21 5 3 4 2 – 41 –1 5 2 2 – 31 3 –1 0 2 = 1 –10 –6 –8 2 – 1 –4 20 8 2 – 1 9 –3 0 2 = 1 –10 – (–4) – 9 –6 – 20 – (–3) –8 – 8 – 0 2 = 1 –15 –23 –162 Example 6 Given vectors = 2 a i + 3j – 5k, b = i – 4j + 2k and c = 4i + 7j – 5k. Find in terms of i, j and k the vector. (a) 3a – b – c (b) 7a – 2b – 4c Solution: (a) 3a – b – c = 3(2i + 3j – 5k) – (i – 4j + 2k) – (4i + 7j – 5k) = (6i – i – 4i) + (9j + 4j – 7j) + (–5k – 2k + 5k) 3a – b – c = i + 6j – 2k (b) 7a – 2b – 4c = 7(2i + 3j – 5k) – 2(i – 4j + 2k) – 4(4i + 7j – 5k) = (14i – 2i – 16i) + (21j + 4j – 28j) + (–35k – 4k + 20k) 7a – 2b – 4c = –4i – 3j – 19k
236 Mathematics Term 1 STPM Chapter 6 Vectors 6 Example 7 Relative to an origin O, points P, Q and R have position vectors given respectively by 1 4 –3 –12 , 1 6 9 7 2 and 1 5 3 3 2 . (a) Express, in terms of i, j and k the vectors →PQ and →PR. (b) Show that the points P, Q and R are collinear. Solution: (a) →OP = 4i – 3j – k, →OQ = 6i + 9j + 7k and →OR = 5i + 3j + 3k →PQ = →OQ – →OP = 6i + 9j + 7k – (4i – 3j – k) →PQ = 2i + 12j + 8k →PR = →OR – →OP = 5i + 3j + 3k – (4i – 3j – k) →PR = i + 6j + 4k (b) →PR = i + 6j + 4k →PR = 1 2 →PQ and P is the common point. \ P, Q and R are collinear. Exercise 6.2 1. If a = 3i – 2j and b = –5i – 6j, express in terms of i and j (a) a + b (b) –3a – 5b (c) 1 2 a – 3 4 b (d) 4b – 5a (e) 3 4 (a – b) 2. Given that u = i + j and v = –i + 3j, express the vector w = 3i – 5j in the form of au + bv where a and b are two scalars. 3. If a = 2i + j – 4k and b = 3i + 2j + k, express in terms of i, j and k, (a) b – a (b) 1 2 (3b – 5a) (c) –3a – 2b 4. A, B, C and D are points (0, 1, 2), (–1, 4, 2), (1, 0, 5) and (–3, 4, –6) respectively. Find the vectors representing (a) →AB, (b) →BD, (c) →AD, (d) →BC. 5. →OP represents a position vector p. Write down the coordinates of p if (a) p = 5i – 6j – k (b) p = 2j – 3k (c) p = –i + 5k (d) p = –4k (e) p = 9j 6. Show that the three vectors p = 3i – 2j + k, q = 2i + j – 4k and r = i – 3j + 5k form the sides of a triangle. 7. Given that p = –i + 2j + 3k, q = 3i + j – 2k and r = i + 5j + 2k. If s = –3i – j + 6k can be expressed in the form of l1 p + l2 q + l3 r, determine the values of l1 , l2 and l3 . 8. Prove that the three vectors 2i + j – 4k, 3i – 2j + k and i – 3j + 5k form the sides of a right-angled triangle.
Mathematics Term 1 STPM Chapter 6 Vectors 6 237 9. If a = 3i – j – 4k, b = 2i + 4j – 3k and c = i + 2j – k, find the unit vector parallel to 3a – 2b + 4c. 10. Calculate the unit vector along the sum of the vectors i + j – k, 2i – 2j + k and 3i + j – 2k. 11. The three points A, B and C have position vectors a, b and c respectively. If c = 3b – 2a, show that the points A, B and C are collinear. Scalar product of two vectors The scalar product (dot product) of two non-zero vectors a and b, written as a · b, is given by a · b = |a||b| cos q, where q is the angle between a and b. b O a Figure 6.13 Note that q is the angle where the two vectors either converges towards point O or diverges from the point O. Case 2: Both vectors diverges from O Case 1: Both vectors converges towards O O Figure 6.14 To find the angle, q between the two vectors, a and b, we use cos q = a · b |a||b| or q = cos–1 1 a · b |a||b| 2 Dot Product of Vectors VIDEO
238 Mathematics Term 1 STPM Chapter 6 Vectors 6 This is valid whether both a and b are two dimensional vectors (on the plane) or both are three dimensional vectors (in the space). Notice that a · b is a scalar. If c is another vector, (a · b) · c has no meaning because we cannot form a scalar product of a scalar (a · b) with the vector c. Note that a · b = 0 if and only if a and b are perpendicular. Properties of the scalar product 1. a · b = b · a 2. a · (b + c) = a · b + a · c Example 8 Find the angle between the vectors 2i + 5j + 3k and 10i – 4j + k. Solution: Let q be the angle between the two given vectors. (2i + 5j + 3k) · (10i – 4j + k) = |2i + 5j + 3k||10i – 4j + k| cos q 20 – 20 + 3 = 22 + 52 + 32 · 102 + (–4)2 + (1)2 · cos q 3 = 38 · 117 · cos q cos q = 3 38 · 117 q = 87°25´ Example 9 Determine the angle between vector a = i + j – k and b = 3i + j + 4k. Solution: Let q be the angle between vector a and b cos q = (i + j – k) · (3i + j + 4k) 12 + 12 + (–1)2 · 32 + 12 + 42 cos q = 3 + 1 – 4 3 · 26 = 0 q = 90° ⇒ a · b = 0 implies that a and b are perpendicular to each other. Scalar product in component form Consider two-dimensional vectors a and b where a = x1 i + y1 j and b = x2 i + y2 j a · b = (x1 i + y1 j) · (x2 i + y2 j) = x1 x2 i · i + x1 y2 i · j + y1 x2 j · i + y1 y2 j · j = x1 x2 + y1 y2 since i · i = j · j = 1 i · j = j · i = 0
Mathematics Term 1 STPM Chapter 6 Vectors 6 239 For three-dimensional vectors p and q where p = x1 i + y1 j + z1 k q = x2 i + y2 j + z2 k p · q = (x1 i + y1 j + z1 k) · (x2 i + y2 j + z2 k) = x1 x2 i · i + x1 y2 i · j + x1 z2 i · k + x2 y1 j · i + y1 y2 j · j + y1 z2 j · k + x2 z1 k · i + y2 z1 k · j + z1 z2 k · k = x1 x2 + y1 y2 + z1 z2 since i · i = j · j = k · k = 1 i · j = j · k = k · i = 0 i, j and k are unit vectors. They are perpendicular to each other, i.e. |i| = |j| = |k| = 1 and i j j k k i So, three unit vectors, i, j and k formed an orthonormal system. k j 1 1 1 i y z x Figure 6.15 From the scalar product of two vectors, a · b = |a||b| cos q i · j = |i||j| cos 90° = 0 j · k = |j||k| cos 90° = 0 k · i = |k||i| cos 90° = 0 \ i · j = j · k = k · i = 0 i · i = |i||i| cos 0° = 1 j · j = |j||j| cos 0° = 1 k · k = |k||k| cos 0° = 1 \ i · i = j · j = k · k = 1 Example 10 Find the scalar product of vectors a = 5i – 2j + k and b = 4i – 6j – 2k. Solution: Scalar product of vectors a and b = a · b = (5i – 2j + k) · (4i – 6j – 2k) = (20i · i – 30i · j – 10i · k – 8j · i + 12j · j + 4j · k + 4k · i – 6k · j – 2k · k = 20(1) – 30(0) – 10(0) – 8(0) + 12(1) + 4(0) + 4(0) – 6(0) – 2(1) = 20 + 12 – 2 since i · i = j · j = k · k = 1 = 30 i · j = j · k = k · i = 0
240 Mathematics Term 1 STPM Chapter 6 Vectors 6 Example 11 Show that the following pairs of vectors are orthogonal. (a) a = 2i – 4j + 2k, b = i + 2j + 3k (b) u = j – 2k, v = 4i – 2j + k Solution: (a) a · b = (2i – 4j + 2k) · (i + 2j + 3k) = 2 – 8 + 6 a · b = 0 a · b = 0 ⇔ a b \ a and b are orthogonal. (b) u · v = (j + 2k) · (4i – 2j + k) = –2 + 2 u · v = 0 u · v = 0 ⇔ u v \ u and v are orthogonal. Example 12 O k j i P Q M V T W S R The diagram shows a cuboid OPQRSTVW, where O is the origin and M is the mid point of VW. Unit vectors i, j and k are parallel to →OP, →OR and →OS respectively. Given that the length of →OP, →OR and →OS are 8 units, 5 units and 7 units respectively. (a) Express each of the vectors →QS and →QM in terms of i, j and k. (b) Determine the acute angle between →QS and →QM, to the nearest 0.1°.
Mathematics Term 1 STPM Chapter 6 Vectors 6 241 Solution: O k j i P Q M V T W S R θ OS = 7k QV = 7k VM = –4i –8i – 5j (a) →QS = –8i – 5j + 7k →QM = –4i + 7k (b) cos q = →QS · →QM u →QSu u →QMu = (–8i – 5j + 7k) · (–4i + 7k) (–8)2 + (–5)2 + (7)2 · (–4)2 + (7)2 cos q = 32 + 49 138 · 65 q = cos–1 1 81 8970 2 \ ∠MQS, q = 31.2° Vector product of two vectors The vector product (cross product) of two non-zero vectors a and b is given by a × b = (|a||b| sin q)n ^ , where q is the angle between a and b and n ^ is a unit vector perpendicular to both a and b. The direction in which the vector product of a and b, may point can be obtained from the right hand rule. n b q a fi b a Figure 6.16
242 Mathematics Term 1 STPM Chapter 6 Vectors 6 We multiply two vectors in the plane a and b together to get a new vector a × b which is perpendicular to the plane. By the right hand rule, we find a unit vector n ^ which is perpendicular to the plane. So, n ^ is the unit normal vector that points the way your right thumb points when your fingers twist through the angle q from a to b. n ^ is a unit vector perpendicular to both a and b. The vector product a × b is also perpendicular to both a and b. So, vector product a × b is a scalar multiple of n ^ . Note that a × b = 0 if and only if a and b are parallel. Properties of the vector product 1. a × b = –b × a 2. a × (b + c) = a × b + a × c Vector product in component form From the vector product of two vectors, a × b = (|a||b| sin q)n ^ i × i = (|i||i| sin 0)n ^ = 0 j × j = (|j||j| sin 0)n ^ = 0 k × k = (|k||k| sin 0)n ^ = 0 \ i × i = j × j = k × k = 0 3 unit vectors, [i, j, k], [j, k, i] and [k, i, j] form a right hand rule where the positive vector product of two unit vectors can be obtained. (i) k j i (ii) k j i Figure 6.17(a) Figure 6.17(b) Multiply clockwise, product is positive Multiply anticlockwise, product is negative i × j = k j × i = –k j × k = i i × k = –j k × i = j k × j = –i Consider three-dimensional vectors a and b where a = x1 i + y1 j + z1 k and b = x2 i + y2 j + z2 k a × b = u i x1 x2 j y1 y2 k z1 z2 u Notice that a × b is a vector and that a × b is defined only when a and b are three-dimensional vectors.
Mathematics Term 1 STPM Chapter 6 Vectors 6 243 Example 13 Given that u = 2i – j + 3k and v = i – 2j + k. Find the unit vector that is perpendicular to both the vectors u and v. Solution: u = 2i – j + 3k v = i – 2j + k u × v = u i 2 1 j –1 –2 k 3 1 u = u –1 –2 3 1 ui – u 2 1 3 1 uj + u 2 1 –1 –2 uk = (–1 + 6)i – (2 – 3)j + (–4 + 1)k u × v = 5i + j – 3k |u × v| = 52 + 12 + (–3)2 = 35 Unit vector that is perpendicular to both u and v = u × v |u × v| = 5i + j – 3k 35 = 5 35 i + 1 35 j – 3 35 k Example 14 Show that vector u = –6i + 3j – 12k is parallel with vector v = 2i – j + 4k. Solution: u × v = u i –6 2 j 3 –1 k –12 4 u = u 3 –1 –12 4 ui – u –6 2 –12 4 uj + u –6 2 3 –1 u k = (12 – 12)i – (–24 + 24)j + (6 – 6)k u × v = 0 u × v = 0 if and only if u and v are parallel. Area of a parallelogram with a and b as the adjacent sides a a b A b B O C h = |a| sin Figure 6.18
244 Mathematics Term 1 STPM Chapter 6 Vectors 6 Area of parallelogram OABC = h|b| = |a| sin q |b| = |a||b| sin q = |a × b| Area of ∆OAC = 1 2 Area of parallelogram OABC = 1 2 |a × b| Example 15 Find the area of the parallelogram whose adjacent sides are 4i – j + 5k and 2i + j – 6k. Solution: Vector area of parallelogram = u i 4 2 j –1 1 k 5 –6 u = (6 – 5)i – (–24 – 10)j + (4 + 2)k = i + 34j + 6k Area of parallelogram = 12 + 342 + 62 = 1193 unit2 Example 16 P(1, –2, 3), Q(2, 1, 8) and R(–4, 1, 0) are three vertices of a triangle. Find (a) →PQ × →PR (b) area of the triangle PQR Solution: →PQ = →OQ – →OP = 2i + j + 8k – (i – 2j + 3k) →PQ = i + 3j + 5k →PR = OR – OP = –4i + j – (i – 2j + 3k) →PR = –5i + 3j – 3k (a) →PQ × →PR = u i 1 –5 j 3 3 k 5 –3 u = (–9 – 15)i – (–3 + 25)j + (3 + 15)k →PQ × →PR = –24i – 22j + 18k Area of ∆PQR = 1 2 | →PQ × →PR| = 1 2 × (–24)2 + (–22)2 + (18)2 = 1 2 1384 unit2 = 18.60 unit2