Chapter 24.2 Ellipse PDF

Complex Numbers 187 Chapter E3 60 5.2 CONTENTS Y 5.2.1 Definition 5.2.2 Standard equation of the Ellipse 5.2.3 Equation of Ellipse in other forms 5.2.4 Parametric form of the Ellipse 5.2.5 Special form of an Ellipse 5.2.6 Position of a point with respect to an Ellipse 5.2.7 Intersection of a line an...

Complex Numbers 187 Chapter E3 60 5.2 CONTENTS Y 5.2.1 Definition 5.2.2 Standard equation of the Ellipse 5.2.3 Equation of Ellipse in other forms 5.2.4 Parametric form of the Ellipse 5.2.5 Special form of an Ellipse 5.2.6 Position of a point with respect to an Ellipse 5.2.7 Intersection of a line and an ellipse 5.2.8 Equation of tangent in different forms 5.2.9 Equation of pair of tangents SS1 = T2 5.2.10 Equations of normal in different forms 5.2.11 Auxiliary circle 5.2.12 Properties of Eccentric angles of the conormal points 5.2.13 Chord of contact ID X U D YG Equation of chord with mid point (x1, y1) Equation of the chord joining two points on an Ellipse ST 5.2.15 U 5.2.14 5.2.16 Pole and Polar 5.2.17 Diameter 5.2.18 Subtangent and Subnormal 5.2.19 Concyclic points 5.2.20 Reflection property of an Ellipse Assignment (Basic and Advance Level) Answer Sheet of Assignment B M Z A C S P M Z S A X B Y The Greeks particularly Archimedes (287-212 B.C.) and Apollonius (200 B.C.) studied conic sections for their own beauty. These curves are important tools for present day exploration of outer space and also for research into behaviour of atomic particles. Kepler was first to declare that the planets of our solar system travel around the sun in elliptic path. The ellipse is also used in many art forms and in the construction of bridges. Due to our knowledge of the ellipse, it is now possible to predict accurately the time and place of solar and lunar eclipses. 60 188 Conic Section : Ellipse 5.2.1 Definition. E3 An ellipse is the locus of a point which moves in such a way that its distance from a fixed point is in constant ratio ( b 4 Where b  5, a  3. So e  1  a2 9 4  1  25 5 b2 The focal distances of a point (x1 , y1 ) on the ellipse are b  ey 1  5  4 y1. 5 5.2.4 Parametric form of the Ellipse. x2 a2  y2 b2  1 , Where b 2  a 2 are given by b  ey 1. So, Required distances Conic Section : Ellipse 193 x2 y2  1 a2 b 2 Then the equation of ellipse in the parametric form will be given by x  a cos  , y  b sin  , where  is the Let the equation of ellipse in standard form will be given by eccentric angle whose value vary from 0    2. Therefore coordinate of any point P on the ellipse will be given by The curve represented by x  3(cos t  sin t), y  4 (cos t  sin t) is (a) Ellipse Solution: (c) x2 y2 x2 y2   2 , which represents ellipse.   (1  sin 2 t)  (1  sin 2 t)  9 16 9 16 The distance of the point '  ' on the ellipse (a) a(e  cos  ) (d) Circle x y  (cos t  sin t),  (cos t  sin t) 3 4 Given, x  3(cos t  sin t), y  4 (cos t  sin t)  Squaring and adding, we get Example: 12 (c) Hyperbola E3 Solution: (a) (b) Parabola [EAMCET 1988; DCE 2000] x2 y2   1 from a focus is a2 b 2 (b) a(e  cos  ) (c) a(1  e cos  ) (d) a(1  2e cos  ) ID Example: 11 60 (a cos  , b sin  ) Focal distance of any point P(x , y ) on the ellipse is equal to SP  a  ex. Here x  cos . Hence, SP  a  ae cos   a(1  e cos  ) U 5.2.5 Special forms of an Ellipse. (1) If the centre of the ellipse is at point (h, k ) and the directions of the axes are parallel to the coordinate axes, then (x  h)2 (y  k )2  1 a2 b2 If we shift the origin at (h, k) without rotating the coordinate axes, then x  X  h and y  Y  k D YG its equation is (lx  my  n)2 (mx  ly  p)2   1 where lx  my  n  0 and mx  ly  P  0 a2 b2 mx  ly  p lx  my  n are perpendicular lines, then we substitute  X,  Y , to put the equation in the standard form. l2  m 2 l2  m 2 (2) If the equation of the curve is The foci of the ellipse 25 (x  1)2  9(y  2)2  225 are U Example: 13 (a) Given ellipse is ST Solution: (d) (1, 2), (6, 1) (b) (1,  2), (1, 6) [MP PET 1998, UPSEAT 1991, 2000] (c) (1,2), (1,6) (d) (1, 2), (1,  6) (x  1)2 (y  2)2 X2 Y 2    1 , where X  x  1 and Y  y  2  1 i.e. 9 9 25 25 Here a 2  25 , b 2  9 [Type : Eccentricity is given by e 2  X2 b2  Y2 a2  1] 4 a 2  b 2 25  9 16   , e 5 25 25 a2 4 Foci are given by Y  ae  5    4 5 X  0  y  2  4  y  2  4  6 or 2 x  1  0  x  1. Hence foci are (–1, –6) or (–1, 2). 5.2.6 Position of a point with respect to an Ellipse. 194 Conic Section : Ellipse Let P(x 1 , y 1 ) be any point and let ellipse as if S 1  x 12 a2  y 12 b2 x2 y2   1 is the equation of an ellipse. The point lies outside, on or inside the a2 b 2  1  0,  0,  0 Y P(inside) X 60 P(outside) P(on) X C Y E3 Example: 14 x2 y2 Let E be the ellipse   1 and C be the circle x 2  y 2  9. Let P and Q be the points (1, 2) and (2, 1) respectively. Then 9 4 (a) Q lies inside C but outside E (b) (c) P lies inside both C and E The given ellipse is x2 y2   1. The value of the expression 9 4 x 2 y2   1 is positive for x  1, y  2 and negative for 9 4 ID Solution: (d) [IIT 1994] Q lies outside both C and E (d) P lies inside C but outside E x  2, y  1. Therefore P lies outside E and Q lies inside E. The value of the expression x 2  y 2  9 is negative for both the points P and Q. Therefore P and Q both lie inside C. Hence P lies inside C but outside E. x2 y2  1 a2 b 2......(i) and the given line be y  mx  c D YG Let the ellipse be U 5.2.7 Intersection of a Line and an Ellipse. Eliminating y from equation (i) and (ii), then......(ii) x 2 (mx  c) 2  1 a2 b2 i.e., (a 2 m 2  b 2 )x 2  2mca 2 x  a 2 (c 2  b 2 )  0 The above equation being a quadratic in x, its discriminant  4 m 2 c 2 a 4  4 a 2 (a 2 m 2  b 2 )(c 2  b 2 )   b 2 (a 2 m 2  b 2 )  c 2  U Hence the line intersects the ellipse in two distinct points if a m  b  c in one point if c 2  a 2 m 2  b 2 and does not intersect if a 2 m 2  b 2  c 2. 2 2 2 2 5.2.8 Equations of Tangent in Different formss ST (1) Point form: The equation of the tangent to the ellipse xx yy x2 y2  2  1 at the point ( x 1 , y 1 ) is 21  21  1 2 a b a b (2) Slope form: If the line y  mx  c touches the ellipse x2 y   1 , then c 2  a 2 m 2  b 2. Hence, the a2 b 2 straight line y  mx  a 2 m 2  b 2 always represents the tangents to the ellipse. Points of contact: Line y  mx  a 2 m 2  b 2 touches the ellipse   a 2m  b2  ,  2 2 2 a 2m 2  b 2  a m b x2 y   1 at a2 b 2     (3) Parametric form: The equation of tangent at any point  (a cos  , b sin  ) is x y cos   sin   1 a b Conic Section : Ellipse 195 Note :  The straight line lx  my  n  0 touches the ellipse x2 y2  2  1 , if a 2 l 2  b 2 m 2  n 2. 2 a b x2 y2  2  1 , if a 2 cos 2   b 2 sin 2   p 2 and that 2 a b  The line x cos   y sin   p touches the ellipse 60  a 2 cos  b 2 sin   . point of contact is  ,  p p    Two tangents can be drawn from a point to an ellipse. The two tangents are real and distinct or coincident or imaginary according as the given point lies outside, on or inside the ellipse. E3  The tangents at the extremities of latus-rectum of an ellipse intersect on the corresponding directrix. Important Tips A circle of radius r is concentric with the ellipse   .   ID  r2  b 2 tan 1  2 2  a r x2 y2   1 , then the common tangent is inclined to the major axis at an angle a2 b 2 The locus of the foot of the perpendicular drawn from centre upon any tangent to the ellipse r 2  a2 cos 2   b 2 sin2  (in polar coordinates) x2 y2   1 is (x 2  y 2 )2  a 2 x 2  b 2 y 2 or a2 b 2 U  x2 y2   1 intercepted between the axes is a 2 y 2  b 2 x 2  4 x 2 y 2. a2 b 2 The locus of the mid points of the portion of the tangents to the ellipse  The product of the perpendiculars from the foci to any tangent of an ellipse is equal to the square of the semi minor axis, and the feet of these perpendiculars lie on the auxiliary circle. Example: 15 D YG  The number of values of ‘c’ such that the straight line y  4 x  c touches the curve (a) 0 Solution: (c) (b) 1 We know that the line y  mx  c touches the curve 2 x2  y 2  1 is 4 (c) 2 [IIT 1998] (d) Infinite 2 x y  2  1 iff c 2  a 2 m 2  b 2 2 b a U Here, a2  4, b 2  1, m  4  c 2  64  1  c   65 Example: 16 On the ellipse 4 x 2  9 y 2  1, the points at which the tangents are parallel to the line 8 x  9 y are 2 1  ,  5 5 ST (a) Solution: (b,d)  2 1  (b)  ,   5 5 (c)  2  1   ,   5 5  Ellipse is 1 1 x2 y2   1  a 2  , b 2 . The equation of its tangent is 4 x x   9 y y   1 4 9 1/4 1/9  m x 2 2 4 x 8  1  x     x   2 y  and 4 x 2  9 y 2  1  4 x 2  9 4 5 9 y 9 When x   2 2 1 1 , then y   and when x   , then y  . 5 5 5 5  2 1   2 1  Hence points are  ,  ,  ,  5 5   5 5 Example: 17  2 1  (d)  ,  5 5  If any tangent to the ellipse x2 y2   1 intercepts equal lengths l on the axes, then l= a2 b 2 [IIT 1999] 196 Conic Section : Ellipse a2  b 2 (a) Solution: (b) (b) a2  b 2 (c) The equation of any tangent to the given ellipse is (a 2  b 2 )2 (d) None of these x y cos   sin   1 a b b   a   ,0  and Q  0, This line meets the coordinate axes at P    cos    sin   a2 b 2 a b a b  cos   and sin    cos 2   sin 2   2  2  l 2  a 2  b 2  l  a2  b 2. l l l cos  sin  l l 60  x2 y2   1 , is 9 5 Example: 18 The area of the quadrilateral formed by the tangents at the end points of latus- rectum to the ellipse Solution: (d) (a) 27/4 sq. units (b) 9 sq. units (c) 27/2 sq. units (d) 27sq. units By symmetry the quadrilateral is a rhombus. So area is four times the area of the right angled triangle formed by the tangents and axes in the 1st quadrant. E3 IIT Screening 2003] 2 5 y  5 Now ae  a2  b 2  ae  2  Tangent (in the first quadrant) at one end of latus rectum  2,  is x . 1 3 9 3 5   U 5.2.9 Equation of Pair of Tangents SS1 = T2. ID 1 9 x y   1. Therefore area  4...3  27 sq. units. 9/2 3 2 2 i.e. Pair of tangents: Let P(x 1 , y 1 ) be any point lying outside the ellipse D YG can be drawn to it from P. x2 y2   1 and let a pair of tangents PA, PB a2 b 2 Then the equation of pair of tangents PA and PB is SS 1  T 2 2 x y  2 1  0 2 a b where S  S1  x 12 2 a xx 1 a 2   y 12 2 1  0 yy 1 1  0 b b2 P(x1,y1) B U T A 2 Director circle: The director circle is the locus of points from which perpendicular tangents are drawn to the ellipse. ST Let P(x 1 , y 1 ) be any point on the locus. Equation of tangents through P(x 1 , y 1 ) is given by SS 1  T 2 2   xx 1 yy 1  x2 y2   x 12 y 12      i.e.,  2  2  1   2  2  1    2  2  1 b b b  a  a  a P(x1,y1) B 90° They are perpendicular, So coeff. of x 2  coeff. of y 2  0 1  1  2  2 b a 2   x 12 y 12 y 12   x 1    1  2   a4  b 4 b2  a   A    0 or x 12  y 12  a 2  b 2   C A B Hence locus of P( x 1 , y1 ) i.e., equation of director circle is x 2  y 2  a 2  b 2 Example: 19 The angle between the pair of tangents drawn from the point (1, 2) to the ellipse 3 x 2  2y 2  5 is (a) tan 1 (12 / 5) (b) 1 tan (6 / 5 ) (c) 1 tan (12 / 5 ) (d) tan 1 (6 / 5) [UPSEAT 2001] Conic Section : Ellipse 197 Solution: (c) The combined equation of the pair of tangents drawn (3 x  2y  5)(3  8  5)  (3 x  4 y  5) 2 2 from (1,2) to the ellipse 3 x 2  2y 2  5 is [using SS 1  T ] 2 2  9 x  24 xy  4 y .......  0 2 2 2 h2  ab ab Where a  9 , h  12 , b  4  tan   12 / 5    tan 1 (12 / 5 ) The locus of the point of intersection of the perpendicular tangents to the ellipse (a) (c) x 2  y 2  13 (d) x 2  y2  5 E3 x2 y2   1. Locus is x 2  y 2  9  4 , i.e. x 2  y 2  13. 9 4 The locus of the middle point of the intercept of the tangents drawn from an external point to the ellipse x 2  2 y 2  2 between the coordinate axes, is [IIT Screening 2004] (a) Solution: (c) x2  y2  4 [Karnataka CET 2003] The locus of point of intersection of two perpendicular tangents drawn on the ellipse is x 2  y 2  a 2  b 2 , which is called “director circle”. Given ellipse is Example: 21 (b) 1 1  1 x 2 2y 2 (b) 1 1  1 4 x 2 2y 2 Let the point of contact be R  ( 2 cos  , sin  ) Equation of tangent AB is 2 (c) 1 1  1 2x 2 4y2 (d) 1 1  1 2x 2 y2 Y cos   y sin   1 B A  ( 2 sec  ,0); B  (0, coses  ) D YG  x ID Solution: (c) x 2  y2  9 x2 y2   1 is 9 4 U Example: 20 60 The angle between the lines given by this equation is tan   R X’ Let the middle point Q of AB be (h, k).  h sec  ,k  2 cosec  1 1 1 1  2 1 , sin    cos    2 2k 2 2h 4k h 2 Thus required locus is Q(h,k) O X A P Y’ 1 1  1 2x 2 4y2 5.2.10 Equations of Normal in Different forms. ST U (1) Point form: The equation of the normal at ( x 1 , y 1 ) to the ellipse a2 x b 2y x2 y2   a2  b 2.   1 is 2 2 x1 y1 a b Tangent P(x1,y1) Normal Q(x2,y2) (2) Parametric form: The equation of the normal to the ellipse x2 y2   1 at (a cos  , b sin  ) is a2 b 2 ax sec   by cosec  a 2  b 2. (3) Slope form: If m is the slope of the normal to the ellipse x2 y2   1 , then the equation of normal is a2 b 2 198 Conic Section : Ellipse m(a 2  b 2 ) a2  b 2m 2   a2  mb 2 The coordinates of the point of contact are  ,  2 2 2 a 2  b 2m 2  a b m Note :  If y  mx  c is the normal of     m 2 (a 2  b 2 ) 2 x2 y2 2 c  , then condition of normality is.   1 (a 2  b 2 m 2 ) a2 b 2 x2 a2  y2 b2  1, if a2 b 2  a2  b 2   l 2 m 2  n 2 2  .   E3  The straight line lx  my  n  0 is a normal to the ellipse 60 y  mx   Four normals can be drawn from a point to an ellipse. Important Tips If S be the focus and G be the point where the normal at P meets the axis of an ellipse, then SG  e.SP , and the tangent and normal at P bisect the external and internal angles between the focal distances of P. ID  Y S C G S T X U X Normal P(x1,y1)     D YG Y Any point P of an ellipse is joined to the extremities of the major axis then the portion of a directrix intercepted by them subtends a right angle at the corresponding focus. With a given point and line as focus and directrix, a series of ellipse can be described. The locus of the extermities of their minor axis is a parabola. The equations to the normals at the end of the latera recta and that each passes through an end of the minor axis, if e 4  e 2  1  0 If two concentric ellipse be such that the foci of one be on the other and if e and e’ be their eccentricities. Then the angl e between their axes is e 2  e 2  1. ee The equation of normal at the point (0, 3) of the ellipse 9 x 2  5 y 2  45 is U cos 1 Example: 22 y 3  0 (b) y3 0 Solution: (d) For Here, (x1, y1 )  (0, 3) and a 2  5 , b 2  9 , Therefore Example: 23 Solution: (a) (c) x-axis (d) y-axis x2 y2 (x  x 1 )a 2 (y  y1 )b 2  2  1 , equation of normal at point (x1 , y1 ) , is  2 x1 y1 b a ST (a) [MP PET 1998] (x  0) (y  3).5 .9 or x  0 i.e., y-axis. 0 3 If the normal at any point P on the ellipse cuts the major and minor axes in G and g respectively and C be the centre of the ellipse, then [Kurukshetra CEE 1998] (a) a 2 (CG )2  b 2 (Cg )2  (a2  b 2 )2 (b) a 2 (CG )2  b 2 (Cg )2  (a2  b 2 )2 (c) a 2 (CG )2  b 2 (Cg )2  (a2  b 2 )2 (d) None of these Let at point (x1 , y1 ) normal will be (x  x1 ) 2 (y  y1 )b 2 a  x1 y1 Conic Section : Ellipse 199 At G, y  0  x  CG  y (b 2  a 2 ) x1 (a 2  b 2 ) and at g, x  0  y  Cg  1 2 2 b a x12 y12   1  a 2 (CG )2  b(Cg )2  (a 2  b 2 )2. a2 b 2 The equation of the normal to the ellipse (a) Solution: (b) x  ey  e 3 a  0 (b) x2 y2   1 at the positive end of the latus-rectum is a2 b 2 x  ey  e 3 a  0 The equation of the normal at (x1 , y1 ) to the given ellipse is (c) x  ey  e 2 a  0 60 Example: 24 b2 a2 x b 2 y   a 2  b 2. Here, x1  ae and y1  a x1 y1 a x b 2y  2  a2e 2 ae b / a [  b 2  a2 (1  e 2 ) ]  E3 So, the equation of the normal at positive end of the latus- rectum is 2 (d) None of these ax  ay  a 2 e 2  x  ey  e 3 a  0 e 5.2.11 Auxiliary Circle. If ID The circle described on the major axis of an ellipse as diameter is called an auxiliary circle of the ellipse. x2 y2   1 is an ellipse, then its auxiliary circle is x 2  y 2  a 2 a2 b 2 x2 y2  1. a2 b 2 Draw PM perpendicular from P on the major axis of the ellipse and produce MP to meet the auxiliary circle in Q. Join CQ. The angle XCQ   is called the eccentric angle of the point P on the ellipse. Y P(x,y)  X U D YG X C M Eccentric angle of a point: Let P be any point on the ellipse x2 Y a 2  y2 b2 1 Note that the angle XCP is not the eccentric angle of point P. 5.2.12 Properties of Eccentric angles of the Co-normal points. (1) The sum of the eccentric angles of the co-normal points on the ellipse x2 y2   1 is equal to odd multiple of . a2 b 2 (2) If  ,    are the eccentric angles of three points on the ellipse, the normals at which are concurrent, then U sin(   )  sin(    )  sin(   )  0. (3)Co-normal points lie on a fixed curve: Let P(x 1 , y 1 ), Q( x 2 , y 2 ) , R(x 3 , y 3 ) and S (x 4 , y 4 ) be co-normal points, ST then PQRS lie on the curve (a 2  b 2 )xy  b 2 kx  a  hy  0 This curve is called Apollonian rectangular hyperbola. Note Q(x2,y2) P(x1,y1) T(h,k) R(x3,y3) S(x4,y4) :  The feet of the normals from any fixed point to the ellipse lie at the intersections of the apollonian rectangular hyperbola with the ellipse. Important Tips 200 Conic Section : Ellipse  x2 The area of the triangle formed by the three points, on the ellipse a 2  y2 b2  1 , whose eccentric angles are  ,  and  is             2ab sin  sin  sin .  2   2   2  x2 y2 e 2 sin 2 , where w is one of the angles between the normals at the points  2  1 is given by 2 cot w  2 a b (1  e 2 ) whose eccentric angles are  and Example: 25  . 2 x2 y2   1 , whose distance from the centre of the ellipse is 2, is 6 2 The eccentric angle of a point on the ellipse (a)  / 4 Solution: (a) 60 The eccentricity of the ellipse (b) 3 / 2 (c) 6 cos 2   2 sin 2   2  6 cos 2   2 sin 2   4  3 cos 2   sin 2   2  2 sin 2   1 ID  1 1  sin        / 4 2 2 The area of the rectangle formed by the perpendiculars from the centre of the ellipse to the tangent and normal at the point-whose eccentric angle is  / 4 , is  sin     a2  b 2   a2  b 2    ab    a2  b 2 (b)  2 2 a b   ab   U (a) (c) 1  a 2  b 2 ab  a 2  b 2   ab   (d)  a b  , The given point is ( a cos  / 4 , b sin  / 4 ) i.e. . 2   2 x y So, the equation of the tangent at this point is   2 a b D YG Solution: (a) (d) 7 / 6 5 / 3 Let  be the eccentric angle of the point P. Then the coordinates of P are ( 6 cos  , 2 sin  ) The centre of the ellipse is at the origin, It is given that OP  2 Example: 26 [WB JEE 1990] E3   p1  length of the perpendicular form (0, 0) on (i) = 1 ab 0 0   2 a b 1/a 1/b 2 U ST Therefore, p2  length of the perpendicular form (0, 0) on (ii)  2 ab a b 2   2 2 ab a  b2 2 2 a2b 2  ( 2a ) 2  ( 2b ) 2  a2  b 2   2 2 2(a 2  b 2 )  a  b a2  b 2   ab  ......(i)  a b  a2 x b2y , Equation of the normal at  is   a 2  b 2  2 ax  2 by  a 2  b 2  2 2 a / 2 b / 2   So, area of the rectangle  p1 p 2   a2  b 2   a2  b 2 .....(ii) a2  b 2 2(a 2  b 2 )   ab   5.2.13 Chord of Contact. If PQ and PR 2 be the tangents through point P( x 1 , y1 ) to the ellipse Y 2 xx 1 yy 1 x y  2  1 or  2  1, then the equation of the chord of contact QR is 2 a2 b a b T  0 at ( x 1 , y 1 ) P(x1,y1) Q C X’ R Y 5.2.14 Equation of Chord with Mid point (x1, y1). X Conic Section : Ellipse 201 The equation of the chord of the ellipse T x2 y2   1, whose mid point be (x 1 , y 1 ) is T  S 1 , where a2 b 2 xx 1 yy 1 x 12 y12 ,   1  0 S   1  0 1 a2 b2 a2 b 2 Q(x2,y2) P (x ,y ) 1 1 5.2.15 Equation of the Chord joining two points on an Ellipse. chord joining these two points is y  b sin   x2 y2   1. Then ,the equation of the a2 b 2 E3 Let P(a cos  , b sin  ) ; Q(a cos  , b sin  ) be any two points of the ellipse 60 R(x3,y3) b sin   b sin  (x  a cos  ) a cos   a cos  x     y         cos    sin    cos   a  2  b  2   2  If the chord joining two points whose eccentric angles are  and  cut the major axis of an :  U Note x2 y2   1 is a2 b 2 ID Thus, the equation of the chord joining two points having eccentric angles  and  on the ellipse   ca. 2 2 ca  If  and  be the eccentric angles of the extremities of a focal chord of an ellipse of eccentricity e, then D YG ellipse at a distance ‘c’ from the centre, then tan tan Example: 27 2 tan  2   1e 0. 1e What will be the equation of that chord of ellipse (a) xy 2 (b) x2 y2   1 which passes from the point (2, 1) and bisected on the point 36 9 [UPSEAT 1999] xy 3 (c) x  2y  1 (d) x  2y  4 Let required chord meets to ellipse on the points P and Q whose coordinates are (x 1 , y 1 ) and (x 2 , y 2 ) respectively U Solution: (d)  tan  Point (2,1) is mid point of chord PQ 1 1 (x 1  x 2 ) or x 1  x 2  4 and 1  (y 1  y 2 ) or y 1  y 2  2 2 2 ST  2 Again points (x 1 , y 1 ) and (x 2 , y 2 ) are situated on ellipse; On subtracting  x 12 y 12 x2 y2   1 and 2  2  1 36 9 36 9 x 22  x 12 y 22  y 12 y  y1 (x  x 1 ) 4 1  2     0 or 2 36 9 x 2  x1 4 (y 2  y1 ) 4  2 2  Gradient of chord PQ  y 2  y1 1  x 2  x1 2 Therefore, required equation of chord PQ is as follows, y  1   Alternative: S 1  T  (If mid point of chord is known) 2 x 1y 2 2 12   1  x  2y  4  1  36 9 36 9 1 (x  2) or x  2 y  4 2 202 Conic Section : Ellipse Example: 28 What will be the equation of the chord of contact of tangents drawn from (3, 2) to the ellipse x 2  4 y 2  9 3x  8y  9 (a) (b) 3 x  8 y  25 (c) 3x  4y  9 3x  8y  9  0 (d) Solution: (a) The required equation is T  0 i.e. , 3 x  4 (2 y )  9  0 or 3 x  8 y  9. Example: 29 A tangent to the ellipse x 2  4 y 2  4 meets the ellipse x 2  2y 2  6 at P and Q. The angle between the tangents at P and Q of the ellipse x 2  2y 2  6 is  3 (c) The given ellipse x 2  4 y 2  4 can be written as x2 y2  1 4 1 x Any tangent to ellipse (i) is cos   y sin   1 2 Equation of PQ, i.e. chord of contact is.....(i) A(h,k) P.....(iii) hx ky  1 6 3.....(iv) h/6 k /3 1    h  3 cos  and k  3 sin  (cos  ) / 2 sin  1 U Since (ii) and (iv) represent the same line,  Q ID Let the tangents at P, Q meet at (h, k ).  6 (d).....(ii) x2 y2  1 6 3 Second ellipse is x 2  2y 2  6 , i.e.  4 60 Solution: (a) (b) E3  2 (a) [IIT 1997] D YG So, h 2  k 2  9 or x 2  y 2  9 is the locus of (h, k ) which is the director circle of the ellipse x2 y2  1 6 3  The angle between the tangents at P and Q will be  / 2. Example: 30 The locus of mid-points of a focal chord of the ellipse (a) Solution: (a) x2 a 2  y2 b  2 ex a (b)  y2 b2   ex a y2 b2  1 is (c)  h2  k2 U hae a 2 a 2 b 2. So, locus of (h, k ) is [EAMCET 1995] x 2  y 2  a2  b 2 hx a 2  ky b 2 (d) None of these  h2 a 2  k2 b2. This passes through (ae, 0) , xe x2 y2  2  2 a a b If  and  are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is cos   cos  cos(   ) ST (a) Solution: (d) a 2 a 2 Let (h, k ) be the mid point of a focal chord. Then its equation is S 1  T or  Example: 31 x2 x2 (b) sin   sin  sin(   ) (c) cos   cos  cos(   ) The equation of a chord joining points having eccentric angles  and  is given by x     y         cos    sin   cos   a  2  b  2   2          If it passes through (ae ,0 ) then e cos    cos   2    2   cos    e  cos    2  2   2 sin    e    2 sin     2  2    cos       cos     2  2   sin   sin    e sin(   )    (d) sin   sin  sin(   ) Conic Section : Ellipse 203 5.2.16 Pole and Polar. Let P(x 1 , y 1 ) be any point inside or outside the ellipse. A chord through P intersects the ellipse at A and B respectively. If tangents to the ellipse at A and B meet at Q(h,k) then locus of Q is called polar of P with respect to ellipse and point P is called pole. Q(h,k) A A A Polar P(x1,y1) Polar B B Pole T B B Q P(x1,y1) Equation of polar: Equation of polar of the point ( x 1 , y 1 ) with respect to ellipse xx 1 a 2  yy 1 b2 1 A x2 y2   1 is given by a2 b 2 E3 Pole 60 Q(h,k) (i.e. T  0 )   a 2l  b 2m   P  , n   n P x2 y2   1 is a2 b 2 U Q ID Coordinates of pole: The pole of the line lx  my  n  0 with respect to ellipse R Note R D YG Q :  The polar of any point on the directrix, passes through the focus.  Any tangent is the polar of its own point of contact. Properties of pole and polar (1) If the polar of P(x 1 , y 1 ) passes through Q(x 2 , y 2 ) , then the polar of Q(x 2 , y 2 ) goes through P(x 1 , y 1 ) and such points are said to be conjugate points. U (2) If the pole of a line l1 x  m 1 y  n1  0 lies on the another line l2 x  m 2 y  n 2  0 , then the pole of the second line will lie on the first and such lines are said to be conjugate lines. (3) Pole of a given line is same as point of intersection of tangents at its extremities. The pole of the straight line x  4 y  4 with respect to ellipse x 2  4 y 2  4 is ST Example: 32 (a) (1, 4) Solution: (b) (b) (1, 1) [EAMCET 2002] (c) (4, 1) Equation of polar of (x 1 , y 1 ) w.r.t the ellipse is xx 1  4 yy 1  4 Comparing with x  4 y  4 (d) (4, 4).....(i).....(ii) x1 4 y1   1  x 1  1, y1  1.  Coordinates of pole (x1 , y1 ) = (1, 1) 1 4 Example: 33 If the polar with respect to y 2  4 ax touches the ellipse x2  y2 (a) 2 (c)  2 x 2   2y 2  1 (4 a2 2 /  2 ) 1 x2 2  y2 2 (b)  1, the locus of its pole is x2 2   2y 2 4 a2 1 (d) None of these [EAMCET 1995] 204 Conic Section : Ellipse Solution: (a) Let P(h, k ) be the pole. Then the equation of the polar is ky  2 a(x  h) or y  This touches x2  2  y2  2 2 2a 2ah. x k k 2  2ah   2a  2 2 2 2  1 , So     2     2 , (using c  a m  b ) k k     x2  2  y2      4 a 2 2   2  1 60  4 a 2 h 2  4 a 2 2  k 2  2. So, locus of (h, k ) is 4 a 2 x 2  4 a 2 2   2 y 2 or 5.2.17 Diameter of the Ellipse. Equation of a diameter to the ellipse x2 y2   1 : Let y  mx  c be a a2 b2 x2 y2   1 , where m is a constant and c is a a2 b 2 ID system of parallel chords of the ellipse E3 Definition : The locus of the mid- point of a system of parallel chords of an ellipse is called a diameter and the chords are called its double ordinates i.e. A line through the centre of an ellipse is called a diameter of the ellipse. The point where the diameter intersects the ellipse is called the vertex of the diameter. variable. U The equation of the diameter bisecting the chords of slope m of the ellipse Y y=mx+c X x2 a 2  y2 b 2 X y 1 Y  b2 a 2m x x 2 y2 b2   1 is y   x , which is a2 b 2 a 2m D YG passing through (0, 0). Conjugate diameter: Two diameters of an ellipse are said to be conjugate diameter if each bisects all chords parallel to the other. Y Conjugate diameter of circle i.e. A A  and B B  are perpendicular to each other. B A Q P Hence, conjugate diameter of ellipse are PP  and Q Q . Hence, angle between 90° conjugate diameters of ellipse  90 o. Now the coordinates of the four extremities of two conjugate diameters are P(a cos  , b sin  ) ; P (a cos  ,b sin  ) ; Q(a sin  , b cos  ); Q (a sin  ,b cos  )  X X C P A B Q Y U If y  m 1 x and y  m 2 x be two conjugate diameters of an ellipse, then b2 a2 (1) Properties of diameters (i) The tangent at the extremity of any diameter is parallel to the chords it bisects or parallel to the conjugate diameter. (ii) The tangent at the ends of any chord meet on the diameter which bisects the chord. (2) Properties of conjugate diameters (i) The eccentric angles of the ends of a pair of conjugate diameters of an ellipse differ by a right angle, ST m1m 2  i.e.      2 (ii) The sum of the squares of any two conjugate semi-diameters of an ellipse is constant and equal to the sum of the squares of the semi axes of the ellipse, i.e. CP 2  CD 2  a 2  b 2 (a cos ' , b sin  ' ) D A P(a cos , b sin  ) A C P D Conic Section : Ellipse 205 (iii) The product of the focal distances of a point on an ellipse is equal to the square of the semi-diameter which is conjugate to the diameter through the point, i.e., SP.S P  CD 2 P(a cos , b sin  ) D S C S D 60 P (iv) The tangents at the extremities of a pair of conjugate diameters form a parallelogram whose area is constant and equal to product of the axes, i.e. Y Area of parallelogram  (2a)(2b) = Area of rectangle contained under major and minor axes. Q M D P E3 R X R C P X D Q Y ID (v) The polar of any point with respect to ellipse is parallel to the diameter to the one on which the point lies. Hence obtain the equation of the chord whose mid point is (x 1, y 1 ) , i.e. chord is T  S 1. (3) Equi-conjugate diameters: Two conjugate diameters are called equi-conjugate, if their lengths are equal i.e. (CP )  (CD)2 U 2  a 2 cos 2   b 2 sin 2   a 2 sin 2   b 2 cos 2  D YG  a 2 (cos 2   sin 2  )  b 2 (cos 2   sin 2  )  0  (a 2  b 2 )(cos 2   sin 2  )  0  (a 2  b 2 )  0 ,  cos 2  0. So,   (CP )  (CD)    4 or 3 4 (a 2  b 2 ) for equi-conjugate diameters. 2 Important Tips If the point of intersection of the ellipses a2 2  b2 2 x2 a 2  y2 b 2  1 and x2  2  y2 2  1 be at the extremities of the conjugate diameters of the former, 2 ST then U   The sum of the squares of the reciprocal of two perpendicular diameters of an ellipse is constant.  In an ellipse, the major axis bisects all chords parallel to the minor axis and vice-versa, therefore major and minor axes of an ellipse are conjugate diameters of the ellipse but they do not satisfy the condition m1. m 2  b 2 / a 2 and are the only perpendicular conjugate diameters. Example: 34 If one end of a diameter of the ellipse 4 x 2  y 2  16 is ( 3 , 2) , then the other end is (a) Solution: (c) ( 3 , 2) (b) ( 3 ,2) (c) ( 3 ,2) (d) (0, 0) Since every diameter of an ellipse passes through the centre and is bisected by it, therefore the coordinates of the other end are ( 3 ,2). Example: 35 If  and  are eccentric angles of the ends of a pair of conjugate diameters of the ellipse x2 a2  y2 b2  1 ,then    is equal to 206 Conic Section : Ellipse (a) Solution: (a)   (b) 2  (c) 0 (d) None of these Let y  m 1 x and y  m 2 x be a pair of conjugate diameter of an ellipse Q (a cos , b sin ) be ends of these two diameters. Then m 1 m 2  a 2  y2 b2  1 and let P(a cos  , b sin  ) and  b2 a2 b sin   0 b sin   0  b 2   2  sin  sin    cos  cos   cos(   )  0       / 2. a cos   0 a cos   0 a 60  x2 5.2.18 Subtangent and Subnormal. E3 Let the tangent and normal at P(x 1 , y 1 ) meet the x-axis at A and B respectively. a2 x2 y2 DA  CA  CD   x1 is   1 x1 a2 b 2 Length of sub-normal at P(x 1, y1 ) to the ellipse x2 y2   1 is a2 b 2 ID Length of subtangent at P(x 1 , y 1 ) to the ellipse X X C B D A Y :  The tangent and normal to any point of an ellipse bisects respectively the internal and external angles D YG Note P(x1,y1) U   b2 b2 BD  CD  CB  x 1   x 1  2 x 1   2 x 1  (1  e 2 )x 1. a   a Y A between the focal radii of that point. Example: 36 5 3  x2 y2 , 2  of the ellipse Length of subtangent and subnormal at the point    1 are  2  25 16   (a) Here a 2  25 , b 2  16 , x 1  U Solution: (a)  5 3 10  8 3  ,   2 5 3   ST Length of subnormal   5 3 10  8 3 ,  (b)   2 3  10  (c)  5 3 12  16 3  ,   2 5 3   (d) None of thee a2 25 5 3 5 3 10 5 3  x1    . Length of subtangent =. 2 x1 2 2 5 3 /2 3 b2 16   5 3  8 3 x   2 1 25  2  5 a 5.2.19 Concyclic points. Any circle intersects an ellipse in two or four points. They are called concyclic points and the sum of their eccentric angles is an even multiple of . Q() P( ) R() S() If  , . ,  be the eccentric angles of the four concyclic points on an ellipse, then         2n  , where n is any integer. Conic Section : Ellipse 207 Note :  The common chords of a circle and an ellipse are equally inclined to the axes of the ellipse. Important Tips The centre of a circle x 2  y 2  2 gx  2 fy  c  0 passing through the three points, on an ellipse  2  2   cos   cos   cos   cos(     ) and  f   b  a  4a    a2 y2  b2  1 (whose eccentric angles are   sin   sin   sin   sin(     )   60  a2  b 2 a,  ,  ) is  g    4a x2 P CP and DCD are conjugate diameters of an ellipse and  is the eccentric angles of P. Then the eccentric angles of the point where the circle through P, P , D again cuts the ellipse is  / 2  3. E3  5.2.20 Reflection property of an Ellipse. (a) Solution: (a) X A  C S Reflected ray N S Normal X A B A ray emanating from the point (3,0) is incident on the ellipse 16 x 2  25 y 2  400 at the point P with ordinate 4. Then the equation of the reflected ray after first reflection is 4 x  3 y  12 (b) 3 x  4 y  12 D YG Example: 37 U ID Let S and S  be the foci and PN the normal at the point P of the ellipse, then SP S   SQ S . Hence if an incoming light ray aimed towards one focus strike the concave side of the mirror in Light B Tangent Q P ray the shape of an ellipse then it will be reflected towards the other focus. (c) 4 x  3 y  12 (d) 3 x  4 y  12 For point P y-coordinate =4 Y (0,4) P Given ellipse is 16 x 2  25 y 2  400 16 x 2  25 (4 ) 2  400 ,  x  0 co-ordinate of P is (0, 4) e2 1 X (–3,0) S’ C 16 9  25 25 3 5 U  e  X  Foci ( ae ,0 ) , i.e. (3,0) ST  Equation of reflected ray (i.e. PS ) is x y   1 or 4 x  3 y  12. 3 4 *** Y S(3,0)