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Chap-12 - Conic Sections

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Chap-12 - Conic Sections

- 1. TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 12 P S Conic Sections Sections of cones, equations of conic sections in standard forms. Condition for y = mx + c to be a tangent and point (s) of tangency Conic sections, namely a point, pair of straight lines, circle, ellipse, parabola and hyperbola are called so because they can be obtained when a cone (or double cone) is cut by a plane. The mathematicians associated with the study of conics were Euclid, Aristarchus and Apollonius. Most of the objects around us and in space have shape of conic- sections. Hence study of these becomes a very important tool for present knowledge and further exploration. Conic-sections as Sections of Right Circular Cone(s) 1. When a double right circular cone is cut by a plane parallel to base at the common vertex, the cutting profile is a point. C H A P T E R CHAPTER INCLUDES : Conic sections General equation of conics Parabola Ellipse Hyperbola Solved examples 2. When a double right circular cone is cut by plane, parallel to its common axis, the cut profile is hyperbola 3. When a right circular cone is cut by a plane parallel to its base the cutting profile is a circle. Syllabus
- 2. JEE main Conic Sections TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 4. When a right circular cone is cut by any plane through its vertex, the cutting profile is a pair of straight lines through its vertex 5. When a right circular cone is cut by a plane parallel to a generator of cone, the cutting profile is a parabola. 6. When a right circular cone is cut by a plane which is neither parallel to any generator / axis nor parallel to base, the cutting profile is an ellipse. Hence a point, a pair of intersecting straight lines, circle, parabola, ellipse and hyperbola, all are conic- sections. All the conic sections are plane or two dimensional curves. The conic section is the locus of a point which moves such that the ratio of its distance from a fixed point (focus) to the distance from a fixed straight line (directrix) is constant (e), e is called eccentricity of conic i.e., PS e PK e = 1 for parabola e < 1 for ellipse e > 1 for hyperbola A line through focus and perpendicular to directrix is called - axis. The vertex of conic is that point where the curve intersects the axis. PS e PS2 e2PK 2 PK K directrix Ax+By+C=0 Ax By C 2 or (x )2 (y )2 e2 A2 B2 P(x, y) S(a, b) focus axis
- 3. Conic Sections JEE main TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 2 y S (a, 0) Note : (i) For parabola PN2 = 4AS.AN. (ii) Two parabolas are said to be equal when their latus recta are equal. Simplification shall lead to the equation of the form ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 So equation of a conic is the general equation of second degree. If = abc + 2fgh – af2 – bg2 – ch2 if 0 a pair of straight lines if 0 and h = 0 and a = b a circle if 0 and h2 = ab a parabola if 0 and h2 < ab an ellipse if 0 and h2 > ab a hyperbola PARABOLA The parabola shown has the equation y2 = 4ax. z For this parabola M (i) Vertex A is (0, 0) (ii) Focus S is (a, 0) (iii) Equation of directrix zz´ is x = – a. (iv) Equation of axis AX is y = 0 K Focal chord : A chord of a parabola which passes through the focus is called as focal chord. Focal distance : The distance, of a point on the parabola, from z the focus is called as focal distance of the point. y P L N x A S(a, 0) L Latus rectum : The double ordinate LSL´ through the focus is the latus rectum. Its equation is x = a and its length is 4a units. The equation of the parabola, whose focus is (3, –4) and directrix is the line y + x = 2 is Solution : PS2 = PM2 Gives x y 2 2 (x 3)2 (y 4)2 or 2x2 + 2y2 + 18 + 32 – 12x + 16y = x2 + y2 + 4 – 4x – 4y + 2xy or x2 + y2 – 2xy – 8x + 20y + 46 = 0 Equations of Parabola in Standard Forms (i) y2 = 4ax Focus : (a, 0) Directrix : x = – a Vertex : (0, 0) Axis : y = 0 x x = –a Illustration 1 :
- 4. JEE main Conic Sections TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 y S y S (–a, 0) x = a y S (0, a) Illustration 2 : (ii) y2 = – 4ax Focus : (–a, 0) Directrix : x = a x Vertex : (0, 0) Axis : y = 0 (iii) x2 = 4ay Focus : (0, a) Directrix : y = – a Vertex : (0, 0) x Axis : x = 0 (iv) x2 = – 4ay Focus : (0, – a) Directrix : y = a x Vertex : (0, 0) Axis : x = 0 Find the focus, vertex, length of latus rectum and directrix of parabola x2 + 4x + 4y + 16 = 0 Solution : (x + 2)2 = –4y – 12 = –4 (y + 3) X2 = –4Y where X = x + 2 and Y = y + 3 or x = (X – 2) and y = (Y – 3) a = 1, hence latus rectum = 4a = 4 units in length. In new coordinates, focus : (0, –1), vertex (0, 0), directrix Y = 1 In original co-ordinates, using x = X – 2 and y = Y – 3 focus (–2, –4), vertex (–2, –3) and directrix y + 3 = 1 or, y + 2 = 0 Parametric co-ordinates The parametric co-ordinates of any point on the parabola y2 = 4ax are given by : x = at2 , y = 2at; t is the parameter Position of a point with respect to the parabola y2 = 4ax. A point P(x1, y1) lies inside, on or outside the parabola y2 = 4ax according as y2 4ax 0,= 0 or > 0 1 1
- 5. Conic Sections JEE main TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 1 1 2 2 1 2 1 1 1 1 Illustration 3 : Equations of a Chord Chord joining (x1, y1) and (x2, y2) on parabola y2 = 4ax is y(y1 + y2) = 4ax + y1y2 Chord joining (at 2 , 2at ) and (at 2 , 2at ) is This chord will be a focal chord is t t = –1 y(t + t ) = 2(x + at t ) 1 2 1 2 Length of chord y = mx + c to the parabola y2 = 4ax Length = Length of focal chord joining (at 2 , 2at ) and (at 2 , 2at ) = (t – t )2 1 1 2 2 2 1 12 Length of focal chord through (at2 , 2at) on y2 = 4ax = at t Chords and Tangents Corresponding to any point P(x1, y1) we define the following expressions for the conic viz, S ax2 2hxy by 2 2gx 2fy c 0 T axx1 h(xy1 x1y) byy1 g(x x1) f (y y1) c S1 ax2 2hx y by2 2gx 2fy c (i) Equation of tangent at P(x1, y1) to y2 = 4ax is T 0 i.e. yy1 = 2a(x + x1) (ii) Equation of tangent in slope form is y mx a m a , 2a Its point of contact is m2 m (iii) The line y = mx + c is a tangent to y2 = 4ax if c a m (iv) If the line y = mx + c intersects the parabola y2 = 4ax then, length of the chord intercepted is x2 x1 where x1 and x2 are abscissa of points of intersection and arise as roots of the equation m2x 2 2x(mc 2a) c2 0 Find the equation of common tangents to the parabolas y2 = 4ax and x2 = 4by. Solution : The general equation of tangent to y2 = 4ax is y mx a m …(1) 2 a Let (1) be a tangent to x2 = 4by ; then, x 4bmx m must have coincident roots 16ab a 1 3 16b2m2 0 ; m m b 1 1 2 2 (1) becomes, b3 y a3 x a3 b3 0 4 m2 1 m2 a(a mc ) 1 m2 1 1
- 6. JEE main Conic Sections TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 1 2 1 1 1 1 1 2 1 1 1 Illustration 4 : (i) Equation of pair of tangents Through any given point P(x1, y1) there pass, in general, two tangents to y2 = 4ax. When P is external to the parabola, the joint equation to the pair of tangents from P is SS1 T 2 (ii) Points of intersection of two tangents The tangents to the parabola S y2 4ax 0 drawn at the points (at2, 2at1) and (at 2, 2at2 ) intersect at the point at1t2, (iii) Chord of contact a(t1 t2 ). Equation of chord of contact of the point P(x1, y1) (i.e. of tangents from P) with respect to the parabola y2 = 4ax is T 0 (iv) Chord with middle point P(x1, y1) has equation : T S1 Find the locus of the mid-point of the chord of the parabola y2 = 4ax, which subtends right angle at the vertex. Solution : Let the mid point of the chord be (x1, y1). Equation of the chord is T = s1. i.e. yy1 = 2a(x + x1) = y 2 – 4ax1..................................................(i) Equation of the parabola is y2 = 4ax........................... (ii) Maping (ii) homogeneous with the help of (i), 2 yy1 – 2a(x x1) y 4ax y 2 – 4ax1 y2 (y 2 – 4ax ) = 4ay xy – 8a2 x2 – 8a2 xx 1 1 1 1 8a2x2 – 4ay1xy + (y 2 – 4ax )y2 + 8a2x x = 0 Q Chords subtends right angle at the vertex (0, 0) a + b = 0 8a2 + y 2 – 4ax1 = 0 y 2 = 4ax1 – 8a2 Required locus y2 = 4a(x – 2a) NORMALS TO THE PARABOLA (i) Equation of normal to the parabola y2 = 4ax at the point P(x , y ) is y y y1 (x x ) 1 1 1 2a 1 (ii) Equation of normal to the parabola y2 = 4ax in slope form is y mx 2am am3 Its foot is given by (am2 , –2am) (iii) Equation of normal to the parabola y2 = 4ax at (at2 , 2at) is y tx 2at at3 (iv) The normals at the points (at2, 2at1) and (at2, 2at2 ) intersect at the point (2a a(t2 t t t2 ), –at t (t + t )) 1 1 2 2 1 2 1 2 (v) Through a given point, in general, three normals can be drawn to a parabola (vi) The sum of the slopes of the normals drawn from a given point to a parabola, is zero. (vii) The sum of the ordinates of the feet of the normals drawn from a given point to a parabola is zero. (viii) If the normal at the point P(at2, 2at ) meets the parabola at Q(at2, 2at ), then t t 2 1 1 2 2 2 1 1 t
- 7. Conic Sections JEE main TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 Illustration 5 : K y K L B 1 L1 M P M Z A S C S N A Z L2 B L2 If three normals to the parabola y2 = x are drawn through a point (c, 0), then show that c 1 . 2 Solution : Any normal to the parabola y2 = 4ax, is y = mx – 2am – am3 . In the parabola, y2 = x, a 1 . 4 Equation of the normal to the parabola y2 = x is y 1 1 3 mx – m – m 2 4 . This passes through (c, 0) 0 1 1 3 cm – m – m 2 4 m(4c – 2 – m2 ) = 0 m = 0 or m2 = 4c – 2 n = 0 or m 2 c – 1 2 1 For the real values of m, c . 2 ELLIPSE An ellipse is the locus of a point which moves such that its distance from a fixed point is e (< 1) times its distance from a fixed straight line. Unlike a parabola an ellipse has (i) two foci (ii) two directrices – one corresponding to each focus. (iii) a center – the point such that all chords through it are bisected at it. (iv) two axes – major and minor axes (v) two vertices (vi) two latus recta Thus, the ellipse shown, has the equation x2 y 2 a2 b2 (i) Center is C(0, 0) (ii) Vertices are A(–a, 0) and A´(a, 0) (iii) Foci are S´(ae, 0) and S(–ae, 0) x (iv) Directrix K´M´Z´ corresponds to focus S´ and has equation x a . e Directrix KMZ corresponds to focus S and has equation x a e 2b2 (v) L1L2 and L1 L2 are double ordinates through the foci and are called latus rectum. Each has length a . (vi) AA´ is the major axis and has length 2a. BB´ is the minor axis and has length 2b where a > b and b2 a2 (1 e2 ) Note : 1. An ellipse is the locus of a point which moves such that the sum of its distances from two fixed points remains constant. In fact, PS + PS´ = 2a for all points P(x, y) on the ellipse. 2. The constant in the above definition should be greater than the distance between the fixed points. This definition is called the physical definition of the ellipse. 1
- 8. JEE main Conic Sections TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 The equation of an ellipse, referred to its axes as the axes of coordinates, with foci (± 2, 0) and latus rectum is 6 units. Solution : ae = 2 centre (0, 0); 2b2 a 6 b2 3a 2 b2 2 2 2 2 e 1 a e a2 a b s 4 a2 3a (a 4)(a 1) 0 a = 4, – 1 a2 3a 4 0 a cannot be negative, hence a = 4 b 2 3 So the equation of ellipse is x2 y 2 1 16 12 Auxillary Circle The circle which is described on the major axis AA´ of an ellipse as diameter is called auxillary circle of the ellipse. Equation of ellipse in parametric form x = a cos , y = b sin ; 0 < 2 y Q B P x A N A B where is the parameter (as shown in the figure above) and is called as eccentric angle corresponding to the point P on the ellipse. The point Q has co-ordinates (a cos , a sin ). Further, PN b . x2 y 2 QN a Position of a point with respect to the ellipse 1 a2 b2 (x)2 (y )2 The point R(x´, y´) lies within, upon or outside the ellipse according as 2 positive. a 1 is negative, zero or b2 The curve represented parametrically by x = t2 + t + 1 and y = t2 – t + 1 is (1) A pair of straight lines (2) An ellipse (3) A circle (4) A parabola Solution : x + y = 2t2 + 2 x – y = 2t x y 2 x y 2 2 2 2x + 2y = x2 + y2 – 2xy + 4 x2 + y2 – 2xy – 2x – 2y + 4 = 0 h = –1, a = 1, b = 1, g = –1, f = –1, c = 4 0 and h2 = ab, so curve is a parabola Illustration 7 : 3x2 4y 2 48 Illustration 6 :
- 9. Conic Sections JEE main TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 Two Standard Forms of the Ellipse Standard Equation x2 y 2 1 (a b) a2 b2 (Horizontal Form of an Ellipse) x2 y 2 1 (a, b) b2 a2 (Vertical Form of an Ellipse) Shape of the Ellipse Centre (0, 0) (0, 0) Equation of major axis y = 0 x = 0 Equation of minor axis x = 0 y = 0 Length of major axis 2a 2a Length of minor axis 2b 2b Foci (± ae, 0) (0, ± ae) Vertices (± a, 0) (0, ± a) Equation of directrices x a e y a e Eccentricity e a2 b2 e a2 b2 a2 a2 Length of latus rectum 2b2 a 2b2 a Ends of latra-recta ae, b2 a b2 a , ae Parametric coordinates (a cos, b sin) (a cos, b sin) Focal radii SP = a – ex1 and S′ P = a + ex1 SP = a – ey1 and S′ P = a + ey1 Sum of focal radii SP + S′P = 2a 2a Distance between foci 2ae 2ae Distance between directrices 2a e 2a e Tangents at the vertices x = ± a y = ± a
- 10. JEE main Conic Sections TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 2 CHORDS AND TANGENTS Corresponding to any point P(x1, y1) we define the following expressions for the conic S ax2 2hxy by 2 2gx 2fy c 0 viz., T axx1 h(xy1 x1y) byy1 g (x x1) f (y y1) c S1 ax2 2hx y by2 2gx 2fy c 1 1 1 1 1 1 x2 y 2 (i) The line y = mx + c intersects the ellipse a2 two distinct points if c 2 a2 m2 b2 two coincident points if c 2 a2 m2 b2 imaginary points if c2 a2 m2 b2 1 at b2 x2 y 2 (ii) Length of the chord intercepted by the ellipse a2 1 on the line y = mx + c is b2 2ab 1 m2 a2m2 b2 c 2 a2m2 b2 x2 y 2 (iii) Equation of tangent at the point (x1, y1) on the ellipse a2 1 is b2 T 0 ; i.e., xx1 yy1 1 a2 b2 (iv) Equation of tangent at (a cos , b sin ) is x cos y sin 1 a b (v) Equation of tangent in slope form is y mx a2m b2 The point of contact of y mx a2m2 b2 is , a2m2 b2 a2 m a2m2 b2 b2 and point of contact of y mx a2m2 b2 is , (vi) Point of intersection of two tangents a2m2 b2 a2m2 b2 a cos 1 2 b sin 1 2 2 2 Tangents at (a cos , b sin ) and (a cos , b sin ) intersect at , 1 1 2 2 cos 1 2 cos 1 2 (vii) Equation of pair of tangents 2 x2 y 2 Through any given point P(x1, y1) there pass, in general, two tangents to the ellipse a2 b2 1. When P is external to the ellipse, the joint equation of the pair of tangents from P is SS1 T 2 a2m2 b2
- 11. Conic Sections JEE main TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 11 16 cos2 11sin2 Find the value of if the tangent at the point 4 cos , sin to the ellipse 16x2 11y2 256 is also a tangent to the circle x2 y 2 2x 15 . Solution : 16x2 11y 2 256 x2 y 2 1 42 2 16 …(1) Equation of tangent at P 4 cos , sin to (1) is (4 cos )x ( sin )y 16 …(2) If (2) is a tangent to the circle x2 y2 2x 15 0 , then, length of the perpendicular from the center (1, 0) to the line (2) is equal to the radius, i.e. 4 units 4 cos 16 4 (cos 4)2 16 cos2 11sin2 5 cos2 11 4 cos2 8 cos 5 0 (2 cos 1) (2 cos 5) 0 cos 1 ; 2 3 Chord of Contact The equation of chord of contact of tangents drawn from P(x1, y1) is T 0 Director Circle It is the locus of point of intersection of tangents which are perpendicular to each other. Its equation is x2 y 2 a2 b2 Equation of Chord in Mid Point Form Equation of the chord with middle point P(x1, y1) is T S1 Illustration 8 : 16 11 16 11 11
- 12. JEE main Conic Sections TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 a b n 2 n Normals to the Ellipse In general, four normals can be drawn from any point to an ellipse and the sum of the eccentric angles of their feet is equal to an odd multiple of two right angles (i) Equation of normal at P(x , y ) is x x1 y y1 1 1 x1 y1 a2 b2 (ii) Equation of normal at (a cos , b sin ) is ax sec by cosec a2 b2 m (b2 a2 ) (iii) Equation of normal in slope form is y mx b2 m2 a2 Equation of chord joining two points The equation of chord joining the points (a cos , b sin ) and (a cos , b sin ) is x cos y sin cos a 2 b 2 2 Show that the straight line lx + my = n, is a normal to the ellipse x2 y 2 1 if a2 b2 2 2 2 . Solution : a2 b2 l 2 m2 x2 y 2 Equation of normal at any point P(a cos , b sin ) to the ellipse 1 a2 b …(1) is ax sec by cosec a2 b2 a sec b cosec …(2) a2 b2 If lx + my = n is also a normal to (1), we must have l m n cos a n l a2 b2 ; sin b m n a2 b2 a 2 b 2 a2 b2 2 l m HYPERBOLA The hyperbola is the locus of a point which moves such that its distance from a fixed point is e (> 1) times its distance from a fixed straight line. As in case of an ellipse, the hyperbola has (i) two foci (ii) two directrices – one corresponding to each focus. (iii) a center – the point such that all chords through it are bisected there at. (iv) two axes – the transverse axis and conjugate axis. (v) two vertices (vi) two latus recta Illustration 9 :
- 13. Conic Sections JEE main TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 y B M P S A Z C Z A S B e e Note : The difference of the focal distances of any point on the hyperbola is equal to length of transverse axis ; thus, PS PS 2a . Illustration 10 : Thus, the hyperbola shown in the adjacent figure has the equation x2 y 2 1 a2 b2 (i) Centre is C(0, 0) (ii) Vertices are A´(–a, 0) and A(a, 0) (iii) Foci are S(ae, 0) ; S´(–ae, 0) x (iv) Directrices are x a e (v) AA´ is the transverse axis and has length 2a ; BB´ is the conjugate axes where B and B´ are two points on the axis of y equidistant f rom C such that CB = B´C = b and where b2 a2(e2 1) . x= – a x= a 2b2 (vi) Length of latus rectum is a . Find the equation of hyperbola with axes as co-ordinate axes and one of its vertex is at a distance 9 and 1 from two foci. Solution : PS – PS = 2a 9 – 1 = 2a a = 4 More over focal distance is a + ae = 9 4 + 4e = 9 5 e = 4 Since, b2 = a2 (e2 – 1) b2 16 25 16 16 = 9 b = 3 x2 y 2 Equation of hyperbola is 16 9 1 Auxillary Circle The circle described on the transverse axis AA´ of a hyperbola as diameter is called as auxillary circle of the hyperbola and its equation is x2 y 2 a2 .
- 14. JEE main Conic Sections TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 2 2 a 1 1 1 1 Equation of Hyperbola in Parametric Form x = a sec ; y = b tan , where 0 2 ( is the parameter) x2 y 2 Position of a Point with respect to Hyperbola 1 a2 b2 (x)2 (y)2 The point R(x´, y´) lies within, upon or outside the hyperbola according as a2 or negative. Equilateral or Rectangular Hyperbola 1 is positive, zero b2 The hyperbola x2 – y2 = a2 , whose asymptotes are at right angles (y = ± x) is called as an equilateral or rectangular hyperbola. If this hyperbola is rotated such that the coordinate axes coincides with the asymptotes, 2 then, its equation reduces to xy = c2 where c . 2 (i) Parametric form of xy = c2 is x = ct; (ii) Centre is (0, 0) (iii) Transverse axis has equation y = x Conjugate axis has equation y = –x (iv) Its eccentricity is e . y y c t x xy = c2 Tangent at ct, c t is t 2y x 2ct Normal at ct, c is xt3 ty ct4 c 0 The eccentricity of hyperbola 9x2 – 16y2 + 72x – 32y – 16 = 0 is Solution : (x 4)2 (y 1)2 16 9 1, modified form of given equation a2 = 16, b2 = 9 2 b2 9 25 5 e 1 a2 1 16 16 so e 4 Chords and Tangents Corresponding to any point P(x1, y1), we define the following expressions for the conic S ax2 2hxy by 2 2gx 2fy c 0 viz., T axx1 h(xy1 yx1) byy1 g (x x1) f (y y1) c S1 ax2 2hx y by2 2gx 2fy c Illustration 11 : 1 1 t
- 15. Conic Sections JEE main TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 x2 y 2 (i) The line y = mx + c intersects the hyperbola a2 real and distinct point if c 2 a2m2 b2 real and coincident points if c2 a2 m2 b2 imaginary points if c2 a2 m2 b2 1 at b2 x2 y 2 (ii) Length of the chord intercepted by the hyperbola a2 1 on the line y = mx + c is b2 x2 y 2 (iii) Equation of tangent at the point (x1, y1) lying on the hyperbola a2 1 is T 0 ; b2 i.e., xx1 yy1 1 a2 b2 (iv) Equation of tangent at (a sec , b tan ) is x sec y tan 1 a b (v) Equation of tangent in slope form is y mx a2m b2 The point of contact of y mx a2m2 b2 is , a2m2 b2 a2m2 b2 a2 m b2 The point of contact of y mx a2m2 b2 is , Point of Intersection of Two Tangents a2m2 b2 a2m2 b2 The tangents at (a sec , b tan ) and (a sec , b tan ) intersect at a sin (2 1) b (cos 1 cos 2 ) sin 2 , sin 1 sin 2 sin 1 Equation to Pair of Tangents x2 y 2 Through any given point P(x1, y1) there pass in general, two tangents to the hyperbola a2 1. When P b2 is external to the hyperbola, the joint equation to the pair of tangents from P is SS1 T 2 x2 y2 x2 y 2 xx yy 2 i.e., 1 1 1 1 1 1 1 a2 b2 2 a2 b2 Find the locus of the mid points of chords of the circle 9x2 16y 2 144 . Solution : x2 y2 16 which are tangents to the hyperbola Let P( ) be any point on the locus ; then, P is the middle point of a chord of the circle x2 y 2 16 …(1) Illustration 12 : 2ab b2 c2 a2m2 1 m2 a2m2 b2 a2m2 b2 a2 b
- 16. JEE main Conic Sections TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 y y b x 2 16 and is also a tangent to the hyperbola 9x2 16y2 144 x2 y 2 1 42 32 …(2) Equation of the chord is x y 2 2 …(3) If (3) is a tangent at (h, k) to (2) then (3) is same as the equation 9hx 16ky 144 …(4) 9h 16k 2 2 144 h 6 2 2 ; k 9 2 2 h2 k 2 2 2 2 2 2 Since (h, k) lies on (2) we have : 16 9 1; i.e., 16 9 ( ) Equation of locus is : 16x2 9y 2 (x2 y 2 )2 Chord of Contact x2 y 2 Equation of chord of contact of the point (x1, y1) with respect to the hyperbola a2 Director Circle 1 is T 0 b2 It is the locus of point of intersection of tangents which are perpendicular to each other. Its equation is x2 y 2 a2 b2 Equation of Chord in Mid Point Form Equation of the chord with middle point (x1, y1) is T S1 Equation of Normal (i) Equation of normal to the hyperbola x2 y 2 a2 b2 1 at the point (x1, y1) is a2 y1 1 2 1 (x x1) (ii) Equation of normal at (a sec , b tan ) is a x cos b y cot a2 b2 SOLVED EXAMPLES Example 1 : Find the vertex, axis, focus, directrix and length of latus rectum of the parabola 2y2 + 3y – 4x – 3 = 0 Solution : Given parabola is 2y2 + 3y – 4x – 3 = 0 or 2y 3 y 4x 3 0 2 or 2y 3 2 4 9 4x 3 0 t
- 17. Conic Sections JEE main TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 3 2 9 3 2 33 or 2y 4 4x 3 0 8 or 2y 4 4x 8 3 2 33 or y 4 2x 32 This is of the form Y2 = 4aX where Y y 3 , 4a = 2, X = 4 x 33 32 Equation of Axis is Y = 0 y 3 0 4 y 3 4 Vertex is (0,0) i.e. X = 0, Y = 0 x 33 , 32 y 3 4 33 , 3 Vertex 32 4 Focus is (a, 0) i.e. X = a, Y = 0 x 33 1 , y 3 0 32 2 4 x 17 , 32 y 3 4 17 , 3 Focus is 32 4 Directrix is X = –a x 33 1 32 2 x 49 (Directrix) 32 Length of latus rectum = 4a = 2 Example 2 : Find the equation of the parabola if the focus is at (6, –6) and vertex is (–2, 2). Solution : Let S and A be the focus and vertex of the parabola. Let axis of parabola meet the directrix at M ( ); then, A is mid point of line segment SM. Hence, + 6 –2 = 2 − 6 2 = 2 = –10 = 10 Thus coordinates of M are (–10, 10). Slope of directrix = 1 slope of axis 1 2 6 2 6 The equation of directrix is given by y – 10 = x + 10 x – y + 20 = 0 ...(1)
- 18. JEE main Conic Sections TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 (x 6)2 (y 6)2 B t1 (0, 0) A C t2 1 2 1 Let P ( x, y) be any point on the parabola. Then x2 – 12x + 36 + y2 + 12y + 36 = x2 y 2 400 40x 40y 2xy 2 x2 + y2 + 2xy – 64x + 64y – 256 = 0 Example 3 : Find the length of the side of an equilateral triangle inscribed in the parabola y2 = 4ax so that one angular point is at the vertex. Solution : Let ABC be the equilateral triangle inscribed in the parabola y2 = 4ax where A(0, 0), B (at 2, 2at1) and C (at 2, 2at2 ) . Since BC is perpendicular to the axis of parabola, we have : t2 = –t1. AB = BC AB2 = BC2 X a2t 4 4a2t 2 16a2t 2 1 1 1 t 4 12t2 1 1 t 2 12 Length of a side = 16a2 12 8 3a Example 4 : The normal at any point P on y2 = 4ax meets the axis in G and the tangent at the vertex in G´. If A be the vertex and the rectangle AGQG´ be completed, show that the equation to the locus of Q is x3 2ax2 y2 Solution : The normal at P(at2 , 2at) is y + tx = 2at + at3 …(1) The coordinates of G and G´ are G(a(2 t 2 ), 0), G(0, at(2 t2 )) Since AGQG´ is a rectangle, the co-ordinates (h, k) of Q are x h a(2 t 2 ) and k at (2 t2 ) k 2 k = t h and h a 2 h h3 2ah2 ak 2 Equation of locus is : x3 2ax2 ay 2 Example 5 : Through the vertex O of the parabola y2 = 4ax, chords OP and OQ are drawn at right angles to one another. Show that for all positions of P, PQ cuts the axis of the parabola at a fixed point. Also find the locus of the middle point of PQ. x y 20 2 y G Q P A G
- 19. Conic Sections JEE main TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 y P O Q i Solution : Let P and Q have coordinates (at2, 2at ) and (at 2, 2at ) respectively 1 1 2 2 Since OP OQ , we have t1t2 = – 4 …(1) Equation of PQ is (t1 t2 )y 2x 8a (x 4a) (t1 t2 ) y 0 x 2 This passes through the point (4a, 0) for all positions of P. Let R(h, k) be any point on the locus ; then, R is the middle point of PQ for some position of P. a(t2 t2 ) h 1 2 ; k a(t1 t2 ) . Also t1t2 4 2 2h k 2 (t1 t2 )2 2t1t2 8 a a k 2 2ah 8a2 0 Equation of locus is : y 2 2ax 8a2 0 Example 6 : Show that the locus of the points such that two of the three normals to the parabola y2 = 4ax from them coincide is 27ay 2 4 (x 2a)2 Solution : Let P(h, k) be any point on the locus. The general equation of normal to y2 = 4ax is y = mx – 2am – am3 …(1) Let the three normals from P have slopes m1, m2 and m3 resp., where m2 = m3 ; then, k mi h 2ami am3 : i 1, 2, 3 m1, m2 and m3 are roots of the equation k mh 2am am3 am3 (2a h)m k 0 m1 m2 m3 0 …(2) m1(m2 m3 ) m2m3 2a h a …(3) m1 m2 m3 k a …(4) Since m2 = m3, we have : m1 2m2 m2 h 2a (by(2)) …(5) (by (3) and (5)) …(6) 2 3a
- 20. JEE main Conic Sections TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 1 3 (x 6)2 (y 7)2 2 m3 k 2a (by (4) and (5)) …(7) h 2a 3 k 2 By (6) and (7) we have 3a 2a Equation of locus is : 4 (x 2a)3 27ay 2 Example 7 : Find the centre, the length of the axes, and the eccentricity of the ellipse 2x2 3y 2 4x 12y 13 0 . Solution : 2x2 3y 2 4x 12y 13 = 0 2(x2 2x) 3(y 2 4y) 13 = 0 2(x 1)2 3(y 2)2 = 1 (x 1)2 (y 2)2 1 2 1 = 1 3 X 2 Y 2 a2 b2 = 1 ; where X = Centre is : X = 0, Y = 0 ; Length of major axes Length of minor axes 2a 2b If e denotes the eccentricity, then, b2 a2(1 e2 ) 1 1 (1 e2 ) ; e 3 2 Example 8 : Find the equation of the ellipse whose focus is the point (6, 7) ; directrix is x + y + 2 = 0 and eccentricity is . Solution : By definition of an ellipse, the focal distance of any point P(x, y) on it is ‘e’ times its distance from the directrix, where ‘e’ denotes the eccentricity. 6 (x 6)2 6 (y 7)2 (x y 2)2 5x2 5y 2 76x 88y 2xy 506 = 0 x – 1, Y = y + 2 , a 1 , 2 b 1 3 x = 1, y = –2 2 2 3 1 3 1 x y 2 3 2
- 21. Conic Sections JEE main TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 5 b2 cos2 a2 sin2 1 d 1 2 Example 9 : Find the equation of the ellipse whose foci are (2, 3), (–2, 3) and whose semi-minor axis is of length . Solution : Let S1(2, 3) and S2(–2, be the two foci and let 2a and 2b denote the lengths of major and minor axes respectively. Then, b ae = 2 b2 a2(1 e2 ) a = 3 and 2ae = S1S2 = 4, where e is the eccentricity of the ellipse. The major axes is y = 3 and center is (0, 3) – the mid point of the foci. Hence, equation of the ellipse is x2 (y 3)2 1 9 5 Example 10 : A tangent to the ellipse x2 4y 2 4 meets the ellipse x2 2y 2 6 at P and Q. Show that the tangents at P and Q on the ellipse x2 2y 2 6 are perpendicular. Solution : Let A(2 cos , sin ) be any point on the ellipse x2 4y 2 4 Equation of tangent at A to (1) is x cos 2 sin y 2 (2) intersects the ellipse x2 2y 2 6 …(1) …(2) …(3) at P and Q ; Let T(h, k) be the point of intersection of the tangents at P and Q to (3) ; then equation of pair of tangents from T to (3) is (x2 2y 2 6) (h2 2k 2 6) (xh 2ky 6)2 x2(k 2 3) y 2(h2 6) 2hk xy 6h x 12ky 3 (h2 2k 2 ) 0 We need to show that h2 k 2 9 Equation of chord of contact of tangents from T to (3) is xh + 2ky = 6 …(4) Since (2) and (4) represent the same line, we have, cos h sin k 3 ; or h2 k 2 9 Let ‘d’ denote the perpendicular distance from the center of the ellipse x2 y 2 to the tangent drawn 1 a2 b2 2 2 b2 at a point P on the ellipse. If F1 and F2 are the two foci of the ellipse, show that (PF1 PF2 ) Solution : 4a 2 x2 y 2 Let P(a cos , b sin ) be the point on the ellipse 1 a2 b …(1) The tangent at which is at a distance ‘d’ from the center of the ellipse. Equation of tangent at P to (1) is b x cos a y sin ab d ab …(2) 3) 5 Example 11 :
- 22. JEE main Conic Sections TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 b 0 d b2 b2 cos2 a2 sin2 d2 a2 b2 d2 (a2 b2 ) cos2 a2 e2 cos2 where e denotes the eccentricity of the ellipse. For any point (x0, y0) on (1), the focal distances from foci F1 and F2 are a + ex0 and a – ex0 respectively. Hence, (PF1 PF2 )2 (2e acos)2 4a2 e2 cos2 (PF1 PF2 )2 4a 2 1 Example 12 : The tangent and normal to the ellipse x2 4y 2 4 at a point P on it meet the major axis in Q and R respectively. If QR = 2, find the eccentric angle of P. Solution : x2 4y 2 4 ; a = 2, b = 1 The tangent at P(a cos , b sin ) is ax cos 4by sin 4 x cos 2y sin 2 Equation of normal at P is ax sec by cosec a2 b2 2x sin y cos 3 sin cos Coordinates of Q and R are given by : Q (2 sec , 0)R 3 cos , 2 …(1) …(2) QR = 2 2 sec 3 cos 2 2 4 3 cos2 4 cos Taking ‘+’ sign, 3 cos2 4 cos 4 0 cos 4 8 6 ; cos 2 3 Taking ‘–’ sign 3 cos2 4 cos 4 0 cos cos 2 3 Example 13 : Find the eccentricity, foci, centre, directrices and the lengths of the transverse and conjugate axis of the hyperbola x2 2y 2 2x 8y 1 0 . 2 4 64 6 2 1
- 23. Conic Sections JEE main TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 Solution : x2 2y 2 2x 8y 1 = 0 (x 1)2 2(y 2)2 6 = 0 (y 2)2 (x 1)2 3 Y 2 X 2 6 = 1 A2 B2 = 1 where Y = y – 2 and X = x – 1 ; A and B The center is : X = 0, Y = 0 ; i.e. x = 1, y = 2. Transverse axis is Y axes and conjugate axes is X axes. Length of transverse axis 2 Length of conjugate axis 2 If e denotes the eccentricity, then, B2 A2(e2 1) e2 – 1 = 2; e Coordinates of foci are X = 0, Y = ± Ae i.e. x = 1 and y 2 3 3 i.e. (1, 5) and (1, –1) Equation of directrices are Y A e i.e. y = 3 and y = 1 Example 14 : The equation of one of the directrices of a hyperbola is 2x + y = 1 and the corresponding focus is (1, 2) and eccentricity e focus. . Find the equation of the hyperbola and the co-ordinates of the center and second Solution : Let S be the focus of (1, 2) and PM be the perpendicular distance of a point P(x, y) on the hyperbola from the corresponding directrix. Then, PS = e·PM. (x 1)2 (y 2)2 (2x y 1) 2 3 5 5 (x2 y 2 2x 4y 5) 3 (4x2 y 2 4xy 4x 2y 1) 7x2 2y 2 12xy 2x 14y 22 = 0 is the equation of the hyperbola. Equation of the perpendicular SZ from S to the directrix 2x + y – 1 = 0 is x – 2y + 3 = 0 3 6 3 6 3 3 S(1, 2) Z A A
- 24. JEE main Conic Sections TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 3 5 ( ( 1 , 7 Co-ordinates of Z are . 5 Since the vertices A and A´ divide SZ internally and externally respectively in the ratio are : :1, their coordinates 5 10 7 3 A , 3 1)5 ( 3 1)5 (5 3) 7 3 10 A , 3 1)5 ( 3 1)5 8 , 11 The centre C of the hyperbola is the mid point of AA´ ; therefore C has coordinates 10 10 . If S0 denotes the coordinates of the second focus, then C is the mid point of SS0. 13 , 1 Coordinates of S are 5 5 3 0