Chapter 24.1 Conic Section PDF

60 148 Conic Section : General 5.0.1. Introduction. Directrix ID or SP  constant = e (eccentricity) PM SP= e. PM E3 The curves obtained by intersection of a plane and a double cone in different orientation are called conic section. In other words “Graph of a quadratic equation (in two variables) is...

60 148 Conic Section : General 5.0.1. Introduction. Directrix ID or SP  constant = e (eccentricity) PM SP= e. PM E3 The curves obtained by intersection of a plane and a double cone in different orientation are called conic section. In other words “Graph of a quadratic equation (in two variables) is a “Conic section”. A conic section or conic is the locus of a point P, which moves in such a way that its distance from a fixed point S always bears a constant ratio to its distance from a fixed straight line, all being in the same plane. M P S 5.0.2. Definitions of Various important Terms. U (1) Focus : The fixed point is called the focus of the conic-section. (2) Directrix : The fixed straight line is called the directrix of the conic section. D YG In general, every central conic has four foci, two of which are real and the other two are imaginary. Due to two real foci, every conic has two directrices corresponding to each real focus. (3) Eccentricity : The constant ratio is called the eccentricity of the conic section and is denoted by e. If e  1 , the conic is called Parabola. If e < 1, the conic is called Ellipse. If e  1 , the conic is called Hyperbola. U If e  0 , the conic is called Circle. If e   , the conic is called Pair of the straight lines. ST (4) Axis: The straight line passing through the focus and perpendicular to the directrix is called the axis of the conic section. A conic is always symmetric about its axis. (5) Vertex: The points of intersection of the conic section and the axis are called vertices of conic section. (6) Centre: The point which bisects every chord of the conic passing through it, is called the centre of conic. (7) Latus-rectum: The latus-rectum of a conic is the chord passing through the focus and perpendicular to the axis. (8) Double ordinate: The double ordinate of a conic is a chord perpendicular to the axis. (9) Focal chord: A chord passing through the focus of the conic is called a focal chord. (10) Focal distance: The distance of any point on the conic from the focus is called the focal distance of the point. 5.0.3. General equation of a Conic section when its Focus, Directrix and Eccentricity are given Conic Section : General 149 Let S ( ,  ) be the focus, Ax  By  C  0 be the directrix and e be the eccentricity of a conic. Let P(h, k ) be any point on the conic. Let PM be the perpendicular from P, on the directrix. Then by definition M  Ah  Bk  C    (h   ) 2  (k   ) 2  e 2   2 2   A B  2 Thus the locus of (h, k ) is (x   ) 2  (y   ) 2  e 2 cartesian equation of the conic section which, ( Ax  By  C ) 2 (A 2   2 ) P(x, y) when Z simplified, can be written in E3 2 S ( ,  ) this is the ax  2hxy  by  2 gx  2 fy  c  0 , which is general equation of second degree. 2 Z 2 60 2 Ax+By+C=0 SP  ePM  SP  e PM 2 5.0.4. Recognisation of Conics. The equation of conics is represented by the general equation of second degree......(i) ID ax 2  2hxy  by 2  2 gx  2 fy  c  0 and discriminant of above equation is represented by  , where   abc  2 fgh  af 2  bg 2  ch 2 U Case I: When   0 In this case equation (i) represents the degenerate conic whose nature is given in the following table. Condition 1.   0 and ab  h 2  0` A pair of coincident straight lines 2.   0 and ab  h 2  0 A pair of intersecting straight lines 3.   0 and ab  h 2  0 A point D YG S. No. Nature of conic Case II: When   0 U In this case equation (i) represents the non-degenerate conic whose nature is given in the following table. ST S. No. Condition Nature of conic 1.   0, h  0, a  b A circle 2.   0, ab  h 2  0 A parabola 3.   0, ab  h 2  0 An ellipse 4.   0, ab  h 2  0 A hyperbola 5.   0, ab  h 2  0 and a  b  0 A rectangular hyperbola 5.0.5. Method to find centre of a Conic. the form 150 Conic Section : General Let S  ax 2  2hxy  by 2  2 gx  2 fy  c be the given conic. Find S S  0,  0 for x, y we shall get the required centre (x, y) x y  hf  bg gh  af  (x , y )   ,  2 2  ab  h ab  h  The equation x 2  2 xy  y 2  3 x  2  0 represents (a) A parabola Solution: (a) [UPSEAT 2001] (b) An ellipse (c) A hyperbola Comparing the given equation with ax  2hxy  by  2 gx  2 fy  c  0 2 Here, a = 1, b = 1, h = – 1, g  2 (d) A circle E3 Example: 1 60 Solve S  S ; x y 3 , f = 0, c = 2 2 Now   abc  2 fgh  af 2  bg 2  ch 2 2 So given equation represents a parabola. ID 9 3 3 i.e.,   0 and h 2  ab  1  1  0 i.e., h 2  ab    (1) (1) (2)  2  (0) (1)  (1) (0)2  1   2(1)2    4 2 2 The centre of 14 x 2  4 xy  11 y 2  44 x  58 y  71  0 is Example: 2 Solution: (a) (b) (2, –3) (c) (–2, 3) U (a) (2, 3)  hf  bg gh  af  , Centre of conic is    ab  h 2 ab  h 2  D YG Here, a  14 , h  2 , b  11 , g  22 , f  29 , c  71  (2)(29 )  (11)(22 ) (22 )(2)  (14 )(29 )   , Centre   2  (14 )(11)  (2) 2   (14 )(11)  (2) ST U Centre  (2, 3). *** [BIT Ranchi1986] (d) (–2, –3) 60 Conic Section : Parabola 151 5.1.1 Definition. A parabola is the locus of a point which moves in a plane such that its distance from a fixed point (i.e., focus) in the plane is always equal to its distance from a fixed straight line (i.e., directrix) in the same plane. ( e  1) SP  PM E3 General equation of a parabola : Let S be the focus, ZZ' be the directrix and let P be any point on the parabola. Then by definition, Z P(x,y)  A2  B2  ID (x   )2  (y   )2  Ax+By+C=0 Directrix M Ax  By  C ( A 2  B 2 ) (x   )2  (y   )2  ( Ax  By  C)2 S(,) Focus Z U Or The equation of parabola whose focus is (5, 3) and directrix is 3 x  4 y  1  0 , is (a) (c) Solution: (a) (3 x  4 y) 2  142 x  256 y  849  0  3x  4y  1   PM 2  PS 2  (x  5) 2  (y  3) 2    9  16   [MP PET 2002] (b) (4 x  3 y) 2  256 x  142 y  849  0 (4 x  3 y) 2  256 x  142 y  849  0 D YG Example: 1 (d) (3 x  4 y) 2  256 x  142 y  849  0 2 Y M P(x,y)  25 (x 2  25  10 x  y 2  9  6 y)  9 x 2  16 y 2  1  12 xy  6 x  8 y  12 xy X'  16 x 2  9 y 2  256 x  142 y  24 xy  849  0 X S(5, 3) O U Y'  (4 x  3 y) 2  256 x  142 y  849  0 ST 5.1.2 Standard equation of the Parabola. Let S be the focus ZZ ' be the directrix of the parabola and (x , y ) be any point on parabola. Let AS  AK  a( 0) then coordinate of S is (a, 0) and the equation of KZ is x  a or x  a  0 Now SP  PM  (SP ) 2  (PM ) 2 Z Y  (x  a) 2  (y  0) 2  (a  x ) 2 which is the equation of the parabola in its standard form. X Directrix  y 2  4 ax L P(x,y ) A S (a,0) K L Z Y P y2=4ax X 152 Conic Section : Parabola Some terms related to parabola Y (a, 2a) Q F(h,2 ah ) L Focal chord P x=a Double ordinate Focal distance Vertex A x+a=0 Z Focus S(a,0) X Axi s Latus rectum F For the parabola y  4 ax , L (a,–2a) Q ' (h,2 ah ) E3 Y 2 60 M Directrix (1) Axis : A straight line passes through the focus and perpendicular to the directrix is called the axis of parabola. ID For the parabola y 2  4 ax, x-axis is the axis. Here all powers of y are even in y 2  4 ax. Hence parabola y 2  4 ax is symmetrical about x-axis. (2) Vertex : The point of intersection of a parabola and its axis is called the vertex of the parabola. The vertex is the middle point of the focus and the point of intersection of axis and the directrix. For the parabola y 2  4 ax, A(0,0) i.e., the origin is the vertex. U (3) Double-ordinate : The chord which is perpendicular to the axis of parabola or parallel to directrix is called double ordinate of the parabola. Let QQ' be the double-ordinate. If abscissa of Q is h then ordinate of Q, y 2  4 ah or y  2 ah (for I Quadrant) and ordinate of Q ' is y  2 ah (for IV Quadrant). Hence coordinates of Q and Q' are (h, 2 ah ) and st D YG th (h,  2 ah) respectively. (4) Latus-rectum : If the double-ordinate passes through the focus of the parabola, then it is called latus-rectum of the parabola. Coordinates of the extremeties of the latus rectum are L(a, 2a) and L' (a,  2 a) respectively. Since LS  L' S  2a  Length of latus rectum LL '  2(LS )  2(L' S )  4 a. ST U (5) Focal Chord : A chord of a parabola which is passing through the focus is called a focal chord of the parabola. Here PP' and LL' are the focal chords. (6) Focal distance (Focal length) : The focal distance of any point P on the parabola is its distance from the focus S i.e., SP. Here, Focal distance SP = PM = x  a Note :  If length of any double ordinate of parabola y 2  4 ax is 2l, then coordinates of end points of this  l2  double ordinate are  , l  and  4a   l2     4 a ,l .   Important Tips 1 ( y1 ~ y 2 )( y 2 ~ y 3 )( y 3 ~ y1 ), where y 1 , y 2 y 3 are the ordinate of the vertices 8a  The area of the triangle inscribed in the parabola y 2  4 ax is  The length of the side of an equilateral triangle inscribed in the parabola y 2  4 ax is 8a 3 (one angular point is at the vertex). Example: 2 The point on the parabola y 2  18 x , for which the ordinate is three times the abscissa, is [MP PET 2003] Conic Section : Parabola 153 (b) (–2, –6) (a) (6, 2) (c) (3, 18) Solution: (d) Given y  3 x , then (3 x )  18 x  9 x  18 x  x  2 and y  6. Example: 3 The equation of the directrix of parabola 5 y 2  4 x is (a) 4x 1  0 (d) (b) [UPSEAT 1998] 4x 1  0 5x  1  0 (c) (d) 5 x  1  0 Example: 4 The point on the parabola y 2  8 x. Whose distance from the focus is 8, has x-coordinate as (b) 2 60 Solution: (c) 4 1 1 The given parabola is y 2  x. Here a . Directrix is x  a   5x  1  0 5 5 5 (a) 0 (2, 6) 2 2 (c) 4 (d) 6 If P( x1 , y1 ) is a point on the parabola y  4 ax and S is its focus, then SP  x 1  a 2 Solution: (d) E3 Here 4 a  8  a  2; SP  8  8  x1  2  x1  6 If the parabola y 2  4 ax passes through (–3, 2), then length of its latus rectum is Example: 5 (a) 2/3 (b) 1/3 [Rajasthan PET 1986, 95] (c) 4/3 (d) 4 ID 1 4 The point (3, 2) will satisfy the equation y 2  4 ax  4  12 a  Latus rectum = 4 | a |  4 |  |  3 3 Solution: (c) 5.1.3 Some other standard forms of Parabola a (1) Parabola opening to left (2) Parabola opening upwards (3) Parabola opening down wards (i.e. y 2  4 ax ); (i.e. x 2  4 ay) ; (i.e. x 2  4 ay ); U (a > 0) P S(–a,0) L M A x-a=0 X D YG Y L Z X Y X U y+a=0 ST Important terms Y Y x  a Equation of the axis y 0 (0,2 a) and y  2a (b) (0, 2a) and y  2 a L S(0, -a) Y x 2  4 ay x 2  4 ay x a (0, 0) (0, a) y  a (0, 0) (0, –a) y=a y 0 x 0 x 0 4a 4a ya 4a ay ax Focus and directrix of the parabola x 2  8 ay are (a) X P Z M (0, 0) (–a, 0) Example: 6 y–a=0 A X L y 2  4 ax x a M X A (0, 0) (a, 0) Focal distance of a point P(x, y) L y 2  4 ax 4a Y P Coordinates of vertex Coordinates of focus Equation of the directrix Length of the latusrectum (a > 0) Z S(0, a) L (a>0) [Rajasthan PET 2001] (c) (2a, 0) and x  2a (d) (2a, 0) and x  2a 154 Conic Section : Parabola Solution: (a) Given equation is x   8 ay Comparing the given equation with x 2  4 AY , A  2a Focus of parabola (0,–A) i.e. (0, –2a) Directrix y  A , i.e. y  2a The equation of the parabola with its vertex at the origin, axis on the y-axis and passing through the point (6, –3) is Example: 7 (a) Solution: (c) y  12 x  6 (b) x  12 y 2 x  12 y 2 (c) Since the axis of parabola is y-axis with its vertex at origin.  Equation of parabola is x 2  12 y. 5.1.4 Special form of Parabola (y – k)2 = 4a(x – h)=a (d) y  12 x  6 2 ; 36= –12a  a  3 E3 Equation of parabola x 2  4 ay. Since it passes through (6, –3) 60 [MP PET 2001] 2 The equation of a parabola with its vertex at (h, k) and axis as parallel to x-axis is (y  k )2  4 a(x  h) Y ID If the vertex of the parabola is ( p , q ) and its axis is parallel to y-axis, then the equation of the parabola is (x  p)2  4 b(y  q) x=h–a Z A When origin is shifted at A ' (h, k ) without changing the direction of axes, its D YG U equation becomes (y  k )2  4 a(x  h) or (x  p)2  4 b(y  q) Equation of Parabola Axis X Directrix X X A Y Focus Directrix Equation of L.R. Length of L.R. (y  K ) 2  4 a(x  h) (h, k ) y k (h  a, k ) x ah  0 x ah 4a (x  p) 2  4 b(y  q) ( p, q ) xp ( p, b  q ) y b q  0 y bq 4b U  Vertex a (h,k) S Important Tips y 2  4 a(x  a) is the equation of the parabola whose focus is the origin and the axis is x-axis. y 2  4 a(x  a) is the equation of parabola whose axis is x-axis and y-axis is directrix.  x 2  4 a(y  a) is the equation of parabola whose focus is the origin and the axis is y-axis.  x 2  4 a(y  a) is the equation of parabola whose axis is y-axis and the directrix is x-axis.  The equation to the parabola whose vertex and focus are on x-axis at a distance a and a' respectively from the origin is y 2  4(a'a)(x  a).  The equation of parabola whose axis is parallel to x-axis is x  Ay 2  By  C and y  Ax 2  Bx  C is a parabola with its axis parallel to yaxis. ST  Example: 8 Solution: (a) Vertex of the parabola x 2  4 x  2y  7  0 is (a) (2,11 / 2) (b) (2,2) [MP PET 1990] (c) (2,11 ) (d) (2,11) 11  11    Equation of the parabola is (x  2) 2  2y  7  4  (x  2) 2  2 y  . Hence vertex is   2, . 2 2     Conic Section : Parabola 155 The focus of the parabola 4 y 2  6 x  4 y  5 is Example: 9 (a) Solution: (b) [Rajasthan PET 1997] (d) (6 / 8,  1 / 2) (b) (5 / 8, 1 / 2) (c) (1 / 2, 5 / 8) Given equation of parabola when written in standard form, we get (8 / 5,2) 2 2 1 3 1 1 3   4  y    6(x  1)   y    (x  1)  Y 2  X where, Y  y  , X  x  1 2 2 2 2 2   1 , x  X 1 2....(i) Focus  X  a, Y  0 ; (b) x 1 (c) [IIT Screening 2001] x 3 2 (d) x  3 2 ID The line x  1  0 is the directrix of the parabola y 2  kx  8  0. Then one of the values of k is 1 8 (b) 8 The parabola is y 2  4 (c) 4 (d) [IIT Screening 2000] 1 4 k 8 8 k  x  . Putting y  Y , x   X. The equation is Y 2  4.. X 4 k 4 k 8 k 8 k k  0 i.e., x    0. But x  1  0 is the directrix. So   1  k  8, 4. 4 k 4 k 4 Equation of the parabola with its vertex at (1, 1)and focus (3, 1)is [Karnataka CET 2001, 2002] D YG (a) Solution: (c) 3 1 1 5  5 1  Focus =   ,  1   ; y  0   8 2 2 8  8 2 1 1 3  x  X. Then the parabola is Y 2  4 X.  The directrix is X  1  0 or  x  1  0 ,  x  2 2 2  The directrix is X  Example: 12  x 1  Here, y 2  4 y  4  4 x  2  0 or (y  2) 2  4  x   2  (a) Solution: (c) 3 8 E3 x  1 Let y  2  Y , Example: 11  a The equation of the directrix of the parabola y 2  4 y  4 x  2  0 is (a) Solution: (d) 3 2 U Example: 10  4a  60 y  Y  (x  1) 2  8(y  1) (b) (y  1) 2  8(x  3) (c) (y  1) 2  8(x  1) (d) (x  3) 2  8(y  1) Given vertex of parabola (h, k )  (1,1) and its focus (a  h, k )  (3,1) or a  h  3 or a  2. We know that as the ycoordinates of vertex and focus are same, therefore axis of parabola is parallel to x-axis. Thus equation of the parabola is (y  k )2  4 a(x  h) or (y  1)  4  2 ( x  1) or (y  1) 2  8(x  1). 2 5.1.5 Parametric equations of a Parabola a The simplest and the best form of representing the coordinates of a point on the parabola y 2  4 ax is U (at2 2at) because these coordinates satisfy the equation y 2  4 ax for all values of t. The equations x  at2 , y  2at taken ST together are called the parametric equations of the parabola y 2  4 ax , t being the parameter. The following table gives the parametric coordinates of a point on four standard forms of the parabola and their parametric equation. Parabola y 2  4 ax y 2  4 ax x 2  4 ay x 2  4 ay Parametric Coordinates Parametric Equations (at 2 ,2at) (at 2 ,2at) (2at, at 2 ) ( 2 at,  at 2 ) x  at 2 y  2 at x  at 2 y  2 at x  2at y  at 2 x  2at , y  at 2 Note Example: 13 2 :  The parametric equation of parabola (y  k )2  4 a(x  h) are x  h  at and y  k  2 at x  2  t 2 , y  2t are the parametric equations of the parabola (a) y2  4x (b) y 2  4 x 2 Solution: (d) Here y y  t and x  2  t 2  (x  2)     y 2  4(x  2) 2 2 (c) x 2  4 y (d) y 2  4(x  2) 156 Conic Section : Parabola 5.1.6 Position of a point and a Line with respect to a Parabola a (1) Position of a point with respect to a parabola: The point P(x 1 , y 1 ) lies outside on or inside the parabola y 2  4 ax according as (2) Intersection of a line and a parabola: Let the parabola be y 2  4 ax And the given line be y  mx  c P P (Outside) (on) P(inside) X 60 y12  4 ax1 , , or  0 Y.....(i).....(ii) Solution: (d) The equation of a parabola is y 2  4 x. P (1,3 ) and Q(1, 1) are two points in the xy-plane. Then, for the parabola (a) P and Q are exterior points (b) P is an interior point while Q is an exterior point (c) P and Q are interior points (d) P is an exterior point while Q is an interior point Here, S  y 2  4 x  0 ; S (1, 3)  3 2  4. 1  0  P(1, 3) is an exterior point. U Example: 14 ID E3 Eliminating y from (i) and (ii) then (mx  c)2 = 4 ax or m 2 x 2  2 x (mc  2a)  c 2  0.....(iii) This equation being quadratic in x, gives two values of x. It shows that every straight line will cut the parabola in two points, may be real, coincident or imaginary, according as discriminate of (iii) >, = or < 0 a  The line y  mx  c does not intersect, touches or intersect a parabola y 2  4 ax , according as c , ,  m a Condition of tangency : The line y  mx  c touches the parabola, if c  m S (1, 1) = 1 2  4. 1  0  Q(1, 1) is an interior point. The ends of a line segment are P (1,3 ) and Q(1, 1). R is a point on the line segment PQ such that PQ : QR  1 : . If R is an D YG Example: 15 interior point of the parabola y 2  4 x , then (a) Solution: (a)  3  (b)     , 1   5    (0, 1) (c) 1 3  2 5   , (d) None of these  1  3   1  3  R   1,  It is an interior point of y 2  4 x  0 iff   4 0  1     1  2 U 3  1  3    1   5  3    1  3    2   2  0         0  (  1)      0 5 1 1   1 1                3 Therefore,     1. But   0  0    1    (0, 1). 5 5.1.7 Equation of Tangent in Different formsa ST (1) Point Form: The equation of the tangent to the parabola y 2  4 ax at a point (x 1 , y 1 ) is yy 1  2a(x  x 1 ) P(x1,y1) T Q(x2,y2) Equation of tangent of all other standard parabolas at (x1, y1) Equation of parabolas Tangent at (x1, y1) y  4 ax yy 1  2a(x  x1 ) 2 Conic Section : Parabola 157 Note : x 2  4 ay xx 1  2a(y  y1 ) x 2  4 ay xx 1  2a(y  y1 ) The equation of tangent at (x 1 , y 1 ) to a curve can also be obtained by replacing x 2 by xx 1 , y 2 by xy 1  x 1 y x  x1 y  y1 , y by and xy by provided the equation of curve is a polynomial of 2 2 2 second degree in x and y. 60 yy 1 , x by   E3 (2) Parametric form : The equation of the tangent to the parabola y 2  4 ax at at 2 ,2at is ty  x  at 2 Equations of tangent of all other standard parabolas at 't' Equations of parabolas Parametric co-ordinates 't' (at 2 , 2 at) x 2  4 ay (2 at, at 2 ) x 2  4 ay (2at,  at 2 ) ty   x  at 2 ID y 2  4 ax Tangent at 't' tx  y  at 2 tx  y  at 2 D YG U  a 2a   is y  mx  a (3) Slope Form: The equation of a tangent of slope m to the parabola y 2  4 ax at 2 m , m  m   Point of contact in terms of slope (m) y 2  4 ax  a 2a   2,  m m  y  mx  a m c y 2  4 ax 2a   a   2 ,  m   m y  mx  a m c x 2  4 ay (2am, am 2 ) y  mx  am 2 c  am 2 (2am,am 2 ) y  mx  am 2 c  am 2 U Equation of parabolas ST x 2  4 ay Equation of tangent in terms of slope (m) Condition of Tangency a m a m Important Tips  If the straight line lx  my  n  0 touches the parabola y 2  4 ax then l n  am.  If the line x cos   y sin   p touches the parabola y 2  4 ax , then P cos   a sin2   0 and point of contact is (a tan 2 ,2a tan  )  If the line 2 x y   1 touches the parabola y 2  4 a(x  b) , then m 2 (l  b)  al 2  0 l m 5.1.8 Point of intersection of Tangents at any two points on the Parabola a The point of intersection of tangents at two points P(at12 ,2at1 ) and Q(at22 ,2at2 ) on the parabola y 2  4 ax is (at 1 t 2 , a(t1  t 2 )). Y 2 (at1 , 2at1 ) P (at1t2,a(t1+t2)) X R O X Q(at22 ,2at2 ) Y 158 Conic Section : Parabola The locus of the point of intersection of tangents to the parabola y 2  4 ax which meet at an angle  is (x  a)2 tan 2   y 2  4 ax. Director circle: The locus of the point of intersection of perpendicular tangents to a conic is known as its director circle. The director circle of a parabola is its directrix. :  Clearly, x-coordinates of the point of intersection of tangents 60 Note P(at12 ,2at1 ) at P and Q on the parabola is the G.M of the x-coordinate of P and Q and y-coordinate is the A.M. of y-coordinate of P and Q. R [at1t2,a(t1+t2)] x  4 by is 2 1 a3 x 1 b3y 2 2 3 a b3 E3  The equation of the common tangents to the parabola y 2  4 ax and 0 Q(at22 ,2at2 )  The tangents to the parabola y 2  4 ax at P(at12 ,2at1 ) and Q(at22 ,2at2 ) intersect at R. Then the area of 1 2 a (t1  t 2 )3 2 ID triangle PQR is 5.1.9 Equation of Pair of Tangents from a point to a Parabolaa U If y 12  4 ax1  0 , then any point P(x 1 , y 1 ) lies out side the parabola and a pair of tangents PQ, PR can be drawn to it from P Y The combined equation of the pair of the tangents drawn from a point to a Q Note D YG parabola is SS '  T 2 where S  y 2  4 ax; S '  y 12  4 ax1 and T  yy 1  2a(x  x 1 ) X O X (x1,y1)P R Y :  The two tangents can be drawn from a point to a parabola. The two tangent are real and distinct or coincident or imaginary according as the given point lies outside, on or inside the parabola. Important Tips Tangents at the extremities of any focal chord of a parabola meet at right angles on the directrix. Area of the triangle formed by three points on a parabola is twice the area of the triangle formed by the tangents at these points.  If the tangents at the points P and Q on a parabola meet in T, then ST is the geometric mean between SP and SQ, i.e. ST  Tangent at one extremity of the focal chord of a parabola is parallel to the normal at the other extremity.  The angle of intersection of two parabolas y 2  4 ax and x 2  4 by is given by tan 1 ST U   Example: 16  1 4 (b)   1 4 (c) [MP PET 1993; MNR 1977] 4 y  2 x   does not meet the parabola y 2  2 x , If   Example: 17 If the parabola y 2  4 ax passes through the point (1,  2) , then the tangent at this point is Solution: (c) x  y 1  0 (b) x  y 1  0 (d)   1 a 1 1 1     m 2.2 4 4 Solution: (b) (a)  SP.SQ 3 a1 / 3 b 1 / 3 2(a 2 / 3  b 2 / 3 ) The straight line y  2 x   does not meet the parabola y 2  2 x , if (a) 2 (c) x  y 10 [MP PET 1998] (d) x  y  1  0  Parabola passes through the point (1, –2), then 4  4 a  a  1. From yy 1  2a(x  x1 )  2 y  2(x  1) Conic Section : Parabola 159  Required tangent is x  y  1  0 The equation of the tangent to the parabola y 2  16 x , which is perpendicular to the line y  3 x  7 is y  3x  4  0 (a) Solution: (a) (b) A line perpendicular to the given line is 3 y  x    y   Here m   (c) 3 y  x  36  0 (d) 3 y  x  36  0 1  x 3 3  1 , c . If we compare y 2  16 x with y 2  4 ax , then a  4 3 3 a  4      36.  Required equation is x  3 y  36  0. m 3 (1 / 3) Condition for tangency is c  Example: 19 3 y  x  36  0 [MP PET 1998] 60 Example: 18 If the tangent to the parabola y 2  ax makes an angle of 45 o with x-axis, then the point of contact is (c) a Parabola is y 2  ax i.e. y 2  4   x 4 a a a a a a  tan 45 o   1  y 1 . From (i), x 1  , So point is  , . 4 2 2y 1 2y 1 4 2 The line x  y  2  0 touches the parabola y 2  8 x at the point (a) (2, –4) (b) (1, 2 2 ) D YG Solution: (d) ax a 2(a / 4 ) (x )  1  y 1 (x  x 1 )  y  2y 1 2y 1 y1 U Example: 20 a a (d)  ,  4 2.....(i) Let point of contact is (x1, y1).  Equation of tangent is y  y 1  Here, m  a a  ,  2 4 ID Solution: (d) a a (b)  ,  4 4 a a  ,  2 2 (a) E3 [Rajasthan PET 1985, 90, 2003] (c) [Roorkee 1998] (4 ,4 2 ) (d) (2, 4) The line x  y  2  0 i.e. x  y  2 meets parabola y 2  8 x , if  y 2  8(y  2)  8 y  16  y 2  8 y  16  0  (y  4 ) 2  0  y  4 ,4  Roots are equal,  Given line touches the given parabola.  x  4  2  2 , Thus the required point is (2, 4). Example: 21 The equation of the tangent to the parabola at point (a / t 2 ,2a / t) is (a) (b) ty  x  at 2 (c) (d) y  tx  (a / t 2 ) y  tx  at 2 Equation of the tangent to the parabola, y 2  4 ax is yy 1  2a(x  x 1 ) U Solution: (a) ty  xt 2  a [Rajasthan PET 1996]  y. Two tangents are drawn from the point (–2, –1) to the parabola y 2  4 x. If  is the angle between these tangents, then tan  = ST Example: 22 2a y  a  a  y t2x  a   2 a x  2     x  2     ty  t 2 x  a t  t t t  t  t2  (a) 3 Solution: (a) (b) 1/3 (c) 2 (d) 1/2 Equation of pair of tangent from (2,1) to the parabola is given by SS 1  T i.e. (y  4 x )(1  8)  [y(1)  2(x  2)] 2 2 2  9 y 2  36 x  [y  2 x  4 ] 2  9 y 2  36 x  y 2  4 x 2  16  4 xy  16 x  8 y  4 x 2  8 y 2  4 xy  20 x  8 y  16  0  tan   1/3 Example: 23 a If   b origin is (a)  / 4 Solution: (b) 2 4  4 (8 ) 12 2 h 2  ab   3 4 8 4 ab b   a 1/3  3 , then the angle of intersection of the parabola y 2  4 ax and x 2  4 by at a point other than the 2 (b)  / 3 Given parabolas are y 2  4 ax....(i) and x 2  4 by (c)  /2.....(ii) (d) None of these 160 Conic Section : Parabola These meet at the points (0, 0), (4 a1 / 3 b 2 / 3 ,4 a 2 / 3 b1 / 3 ) Tangent to (i) at (4 a 1 / 3 b 2 / 3 ,4 a 2 / 3 b 1 / 3 ) is y.4 a2 / 3b1 / 3  2a(x  4 a2 / 3b1 / 3 ) Slope of the tangent (m 1 )  2a 4a 2 / 3 b1 / 3  a1 / 3 2b 1 / 3 Tangent to (ii) at (4 a 1 / 3 b 2 / 3 ,4 a 2 / 3 b 1 / 3 ) is x.4 a 1 / 3 b 2 / 3  2b(y  4 a 2 / 3 b 1 / 3 ) 2a 1 / 3 60 Slope of the tangent (m 2 )  b1/3 (a) Solution: (c) 3y  3x  1 (b) ID The equation of the common tangent touching the circle (x  3) 2  y 2  9 and the parabola y 2  4 x above the x-axis , is 3 y  (x  3) 3y  x  3 (c) 1 Any tangent to y  4 x is y  mx . It touches the circle if 3  m U Example: 24 E3 a1 / 3 2a1 / 3  1/3 1/3 m  m2 If  is the angle between the two tangents, then  tan   1  2b 1 / 3 b 1 / 3 1  m1m 2 a 2a 1 1/3. 1/3 2b b 3 1 3 1  . .  3 ;    60 o  3 2  a 1 / 3  b 1 / 3 2 3     2 b a 2 2 D YG 1 1  or 9(1  m 2 )   3m   or 2  3 , m m  m   3m  (d) 3 y  (3 x  1) 1 m 1 m2 Y 1 (3,0) O 3 For the common tangent to be above the x-axis, m  [IIT Screening 2001] 3 X 1 3  Common tangent is y  1 x  3  3y  x  3 3 Example: 25 If a  0 and the line 2bx  3 cy  4 d  0 passes through the points of intersection of the parabolas y 2  4 ax and x 2  4 ay then d 2  (3b  2c) 2  0 U (a) Solution: (d) (b) d 2  (3b  2c) 2  0 Given parabolas are y 2  4 ax (c) d 2  (2b  3c) 2  0 (d) d 2  (2b  3 c) 2  0.....(i) and x 2  4 ay.....(ii) 2 ST  x2    4 ax  x 4  64 a 3 x  0  x  0 , 4 a  y  0, 4 a from (i) and (ii)   4 a   So points of intersection are (0 ,0 ) and (4 a,4 a) Given, the line 2bx  3 cy  4 d  0 passes through (0,0) and (4 a,4 a)  d  0  d 2  0 and (2b  3 c) 2  0 ( a  0 ) Therefore d  (2b  3c)  0 2 2 5.1.10 Equations of Normal in Different forms a (1) Point form : The equation of the normal to the parabola y 2  4 ax at a point ( x 1 , y 1 ) is y  y1   P(x1,y1) Tangent Normal y1 (x  x 1 ) 2a Conic Section : Parabola 161 Equation of normals of all other standard parabolas at (x1, y1) Equation of parabolas Normal at (x1, y1) y  y1  x 2  4 ay y1 (x  x 1 ) 2a y  y1   x 2  4 ay y  y1  2a (x  x 1 ) x1 2a (x  x 1 ) x1 60 y 2  4 ax E3 (2) Parametric form: The equation of the normal to the parabola y 2  4 ax at (at 2 ,2at) is y  tx  2at  at 3 Equations of normal of all other standard parabola at 't' Equations of parabolas Parametric co-ordinates y 2  4 ax (at 2 , 2 at) Normals at 't' ID y  tx  2at  at 3 x 2  4 ay x  ty  2at  at 3 (2 at,at 2 ) x  ty  2at  at 3 U x 2  4 ay (2 at, at 2 ) (3) Slope form: The equation of normal of slope m to the parabola y 2  4 ax is y  mx  2am  am 3 at the point D YG (am 2 ,2am). Equations of normal, point of contact, and condition of normality in terms of slope (m) Equations of parabola Point of contact in terms of slope (m) Equations of normal in terms of slope (m) Condition of normality (am 2 ,2am) y  mx  2am  am 3 c  2am  am 3 y 2  4 ax (am 2 , 2 am ) y  mx  2am  am 3 c  2am  am 3 x 2  4 ay  2a a  , 2   m m  y  mx  2a  a   2a  , 2  m m   y  mx  2a  U y 2  4 ax ST x 2  4 ay Note a m 2 a m 2 c  2a  c  2a  a m2 a m2 :  The line lx  my  n  0 is a normal to the parabola y 2  4 ax if al(l 2  2m 2 )  m 2 n  0 5.1.11 Point of intersection of normals at any two points on the Parabolaa If R is the point of intersection then point of intersection of normals at any two points P(at12 ,2at1 ) and Q(at 22 ,2at 2 ) on the parabola y 2  4 ax is R [2a  a(t12  t 22  t1 t 2 ),  at1 t 2 (t1  t 2 )] Y X P(at12 ,2at1 ) R A X Q(at22 ,2at2 ) Y 162 Conic Section : Parabola 5.1.12 Relation between 't1' and 't2' if Normal at 't1' meets the Parabola again at 't2' ' If the normal at the point P(at 12 ,2at1 ) meets the parabola y 2  4 ax again at Q(at 22 ,2at 2 ) , then t 2  t 1  X P(at12 ,2at1 ) 60 Y 2 t1 X A Q(at22 ,2at2 ) E3 Y Important Tips If the normals at points (at12 , 2 at) and (at 22 , 2 at 2 ) on the parabola y 2  4 ax meet on the parabola then t1 t 2  2  If the normal at a point P(at 2 ,2at) to the parabola y 2  4 ax subtends a right angle at the vertex of the parabola then t 2  2.  1  If the normal to a parabola y 2  4 ax , makes an angle  with the axis, then it will cut the curve again at an angle tan 1  tan  . 2   The normal chord to a parabola y 2  4 ax at the point whose ordinate is equal to abscissa subtends a right angle at the focus.  If the normal at two points P and Q of a parabola y 2  4 ax intersect at a third point R on the curve. Then the product of the ordinate of P and U ID  Q is 8a 2. If x  y  k is a normal to the parabola y 2  12 x , then k is D YG Example: 26 (a) 3 Solution: (b) (b) 9 U ST From y  y 1   (b) 4 x  8 y  9a  0 (c) 4x  y  a  0 [Rajasthan PET 1984] (d) 4x  y  a  0 y 1 (x  x 1 ) 2a a  a a 9a 9a  0  4 x  8 y  9a  0  2y  x   x    2 y  x  2a   2a  4 4 4 4 The point on the parabola y 2  8 x at which the normal is parallel to the line x  2 y  5  0 is (a) Solution: (b) t 1 6t  3t 3   1 1 k 63 k 9 k 4 x  8 y  9a  0 y a  Example: 28 –3 a  The equation of normal at the point  , a  to the parabola y 2  4 ax , is 4  (a) Solution: (b) (d) Any normal to the parabola y 2  12 x is y  tx  6 t  3 t 3. It is identical with x  y  k if  t  1 and 1  Example: 27 (c) –9 [IIT Screening 2000] (1 / 2, 2) (b) (1 / 2,2) (c) (2,  1 / 2) (d) (2, 1 / 2) k (x  h) or kx  4 y  kh  4 k  0 Let point be (h, k ). Normal is y  k  4 Gradient  K 1 1 1    k  2. Substituting (h, k ) and k  2 in y 2  8 x , we get h . Hence point is  ,  2  2 4 2 2   1  Trick: Here only point  ,  2  will satisfy the parabola y 2  8 x. 2  Example: 29 The equations of the normal at the ends of the latus rectum of the parabola y 2  4 ax are given by Conic Section : Parabola 163 x 2  y 2  6 ax  9 a 2  0 (b) x 2  y 2  6 ax  6 ay  9 a 2  0 (c) x 2  y 2  6 ay  9 a 2  0 (d) None of these The coordinates of the ends of the latus rectum of the parabola y  4 ax are (a, 2a) and (a,2 a) respectively. 2 y   2a (x  a) using y  y 1  1 (x  x 1 ) 2a 2 a  .....(i) The equation of the normal at (a, 2a) to y 2  4 ax is y  2a  x  y  3a  0 Or Similarly the equation of the normal at (a, –2a) is x  y  3 a  0.....(ii) The combined equation of (i) and (ii) is x 2  y 2  6 ax  9 a 2  0. Example: 30 60 Solution: (a) (a) The locus of the point of intersection of two normals to the parabola x 2  8 y, which are at right angles to each other, is [Roorkee 1997] (b) x 2  2(y  6) (c) Given parabola is x 2  8 y Let (4 t 1 2 t 12 ) and  (d) None of these.....(i) Q(4 t 2 , 2 t 22 ) be Normal at P, Q are y  2 t 12   x 2  2(y  6) two points on the parabola (i) 1 (x  4 t 1 ) t1......(ii) and y  2 t 22   1 (x  4 t 2 ) t2......(iii) ID Solution: (a) x 2  2(y  6) E3 (a)  U 1 1 t t (ii)–(iii) gives 2 t 22  t12  x     x 1 2 ,  x  2t1 t 2 (t 2  t1 ).....(iv) t1 t2  t 2 t1  1 From (ii), y  2 t 12  (2 t 1 t 2 (t 2  t 1 )  4 t 1 )  2 t12  2 t2 (t1  t2 )  4  y  2 t 12  2 t 1 t 2  2 t 22  4.....(v) t1 Since normals (ii) and (iii) are at right angles,   From (iv), x  2(t1  t 2 ) and from (v) y  2 t12  2  2 t22  4  y  2[t 12  t 22  1]  2[(t1  t2 )2  2 t1 t2  1] D YG   1 1        1  t1 t 2  1  t  t   1 2  x2  x2  6  x 2  2(y  6), which is the required locus. y  2[(t 1  t 2 ) 2  2  1]  2[(t1  t 2 )2  3  y  2   3   2  4  5.1.13 Co-normal Pointss The points on the curve at which the normals pass through a common point are called co-normal points. Y Q, R, S are co-normal points. The co- normal points are also called the feet of Q the normals. U If the normal passes through point P(x 1 , y 1 ) which is not on parabola, then y 1  mx 1  2am  am  am  (2a  x 1 )m  y 1  0 3 3 P(x1,y1) X......(i) X 0 R S ST Y Which gives three values of m. Let three values of m are m 1 , m 2 and m 3 , which are the slopes of the normals at Q, R and S respectively, then the coordinates of Q, R and S are (am12 ,2am1 ), (am 22 ,2am 2 ) and (am 32,  2 am 3 ) respectively. These three points are called the feet of the normals.  y1 (2a  x 1 ) and m 1 m 2 m 3  a a In general, three normals can be drawn from a point to a parabola. Now m 1  m 2  m 3  0 , m 1 m 2  m 2 m 3  m 3 m 1  (1) The algebraic sum of the slopes of three concurrent normals is zero. (2) The sum of the ordinates of the co-normal points is zero. (3) The centroid of the triangle formed by the co-normal points lies on the axis of the parabola. 164 Conic Section : Parabola (4) The centroid of a triangle formed by joining the foots of the normal of the parabola lies on its axis and is given  am 2  am 22  am 32 2am 1  2am 2  2am 3   am 12  am 22  am 32  = by  1 , ,0    3 3 3     (5) If three normals drawn to any parabola y 2  4 ax from a given point (h, k) be real, then h  2 a for a  1 , 60 normals drawn to the parabola y 2  4 x from any point (h, k) are real, if h  2. (6) Out of these three at least one is real, as imaginary normals will always occur in pairs. 5.1.14 Circle through Co-normal points s Equation of the circle passing through the three (co-normal) points on the parabola y 2  4 ax , normal at which pass through a given point ( ,  ) ; is x 2  y 2  (2a   )x   E3 y 0 2 (1) The algebraic sum of the ordinates of the four points of intersection of a circle and a parabola is zero. (2) The common chords of a circle and a parabola are in pairs, equally inclined to the axis of parabola. ID (3) The circle through co-normal points passes through the vertex of the parabola. (4) The centroid of four points; in which a circle intersects a parabola, lies on the axis of the parabola. The normals at three points P, Q, R of the parabola y 2  4 ax meet in (h, k), the centroid of triangle PQR lies on (a) x 0 U Example: 31 (b) y = 0 (c) x  a [MP PET 1999] y a (d) Since the centroid of the triangle formed by the co-normal points lies on the axis of the parabola. Example: 32 If two of the three feet of normals drawn from a point to the parabola y 2  4 x be (1, 2) and (1, – 2)then the third foot is (a) Solution: (c) D YG Solution: (b) (2, 2 2 ) (b) (2,  2 2 ) (c) (0, 0) (d) None of these The sum of the ordinates of the foot = y1  y2  y3  0  2  (2)  y3  0  y3  0 5.1.15 Equation of the Chord of contact of Tangents to a Parabolas Let PQ and PR be tangents to the parabola y 2  4 ax drawn from any external point P(x 1 , y 1 ) then QR is called U the ‘Chord of contact’ of the parabola y 2  4 ax. Y The chord of contact of tangents drawn from a point ( x 1 , y 1 ) to the parabola ST y 2  4 ax is yy 1  2a(x  x 1 ) Q Chord of contact X (x1,y1)P O R The equation is same as equation of the tangents at the point ( x 1 , y 1 ). Note : Y'  The chord of contact joining the point of contact of two Q(at12 ,2at1 ) perpendicular tangents always passes through focus.  If tangents are drawn from the point ( x 1 , y 1 ) to the parabola y 2  4 ax, then the length of their chord of contact is P(x1,y1) R(at22 ,2at2 ) X Conic Section : Parabola 165 1 (y 12  4 ax 1 )(y 12  4 a 2 ) | a|  The area of the triangle formed by the tangents drawn from (x 1 , y1 ) to y 2  4 ax and their chord of contact is (y 12  4 ax 1 ) 3 / 2. 2a 60 5.1.16 Equation of the Chord of the Parabola which is bisected at a given point s The equation of the chord at the parabola y 2  4 ax bisected at the point ( x 1 , y 1 ) is given by T  S 1, where T  yy 1  2a(x  x 1 ) and S 1  y 12  4ax 1. i.e., yy 1  2a(x  x 1 )  y 12  4 ax1 E3 Q(x2,y2) P(x1,y1) R(x3,y3) ID 5.1.17 Equation of the Chord joining any two points on the Parabola s Let P(at 12 , 2at 1 ), Q(at 22, , 2at 2 ) be any two points on the parabola y 2  4 ax. Then, the equation of the chord joining these points is, y  2at1  2at 2  2at1 at22  at12 (x  at12 ) or y  2at1    2 x  at12 or y(t1  t 2 )  2 x  2at 1 t 2 t1  t 2 D YG U (1) Condition for the chord joining points having parameters t1and t2 to be a focal chord: If the chord joining points (at12 ,2at1 ) and (at 22 ,2at 2 ) on the parabola passes through its focus, then (a,0 ) satisfies the 1 equation y(t1  t 2 )  2 x  2at 1 t 2  0  2a  2at 1 t 2  t1 t 2  1 or t 2   t1 (2) Length of the focal chord: The length of a focal chord having parameters t 1 and t 2 for its end points is a(t 2  t1 )2. Note:  If one extremity of a focal chord is (at12 ,2at1 ) ,then the other extremity (at 22 ,2at 2 ) becomes  a  2a   ,   t 2 t  by virtue of relation t1 t 2  1. 1   1 U a   If one end of the focal chord of parabola is (at 2 ,2at) ,then other end will be  2 ,  2at  and length of t  2 ST  1 chord  a t  . t   The length of the chord joining two point on the parabola y 2  4 ax is between the parabola y 2  4 ax is ' t 1 ' and ' t 2 ' a(t1  t 2 ) (t1  t 2 )2  4  The length of intercept 4 a(1  m 2 )(a  mc ). m2 made by line y  mx  c Important Tips   The focal chord of parabola y 2  4 ax making an angle  with the x-axis is of length 4 a cos ec 2. The length of a focal chord of a parabola varies inversely as the square of its distance from the vertex. 166 Conic Section : Parabola 4 l1 l 2 l1  l 2  If l1 and l2 are the length of segments of a focal chord of a parabola, then its latus-rectum is  The semi latus rectum of the parabola y 2  4 ax is the harmonic mean between the segments of any focal chord of the parabola. Example: 33 60 Solution: (b) If the points (au 2 ,2au) and (av 2 ,2av) are the extremities of a focal chord of the parabola y 2  4 ax , then [MP PET 1998, 93] (a) uv  1  0 (b) uv  1  0 (c) u  v  0 (d) u  v  0 Equation of focal chord for the parabola y 2  4 ax passes through the point (au 2 ,2au) and (av 2 ,2av) 2av  2au 2 2a(v  u) (x  au 2 ) (x  au 2 )  y  2au   y  2au  2 (x  au 2 )  y  2 au  2 v  u av  au a(v  u)(v  u) It this is focal chord, so it would passes through focus (a, 0) 2  0  2au  (a  au 2 )   uv  u 2  1  u 2 ,  uv  1  0 v u  ID Solution: (c) The locus of the midpoint of the line segment joining the focus to a moving point on the parabola y 2  4 ax is another parabola with the directrix [IIT Screening 2002] a a (a) x  a (b) x   (c) x  0 (d) x  2 2 Let M(,  ) be the mid point of PS. at 2  a 2at  0  2  at 2  a, at   ,  2 2  2  a. 2 a2  a or 2a   2  a 2  The locus is y 2  Y (at2,2at) P S(a,0) M( ,) X X O U Example: 34 E3 Trick : Given points (au 2 , 2 au ) and (av 2 , 2av) , then t1  u and t2  v , we know that t1t2  1. Hence uv  1  0. 4a a a  (x  )  4 b(x  b), b   2 2 2  Y D YG  Directrix is (x  b)  b  0 or x  0. Example: 35 The length of chord of contact of the tangents drawn from the point (2, 5) to the parabola y 2  8 x , is 1 3 41 41 (a) (b) (c) (d) 2 41 41 2 2 Solution: (c) Equation of chord of contact of tangents drawn from a point (x1 , y1 ) to parabola y 2  4 ax is yy 1  2a(x  x1 ). So that 5 y  2  2(x  2)  5 y  4 x  8. [MNR 1976] 3 1  Point of intersection of chord of contact with parabola y 2  8 x are  , 2  ,(8, 8), So the length of chord is 2 2  If b, k are the intercept of a focal chord of the parabola y 2  4 ax , then K is equal to ab b a (a) (b) (c) b a b a b a Let ' t1 ' ' t2 ' be the ends of focal chords U Example: 36 Solution: (a) 41. [Rajasthan PET 1999] (d) ab ab ST . t1t2  1. If S is the focus and P, Q are the ends of the focal chord, then   SP  (at12  a)2  (2at1  0)2  a t12  1  b   SQ  a t22  1 a(t 2  1)  12 k t1  ( at12 ,2at1 ) P (Given).... (i)  1  1 1  t 22  2   a 2  1  (Given)  t 2   t1 t1    t1  b....(ii),   t12 k b k S(a,0) Q [Divide (i) by (ii)] (at22 , 2at2 ) ab ab b  a b  k  Putting in (1), we get a  1   b  k b a k  5.1.18 Diameter of a Parabola. The locus of the middle points of a system of parallel chords is called a diameter and in case of a parabola this diameter is shown to be a straight line which is parallel to the axis of the parabola. Y X P(x1,y1) y=mx+c P(h,k) A Y Diameter Q(x2,y2) X Conic Section : Parabola 167 The equation of the diameter bisecting chords of the parabola y 2  4 ax of slope m is y  :    Example: 37 The tangent at the end point of a diameter is parallel to corresponding system of parallel chords. The tangents at the ends of any chord meet on the diameter which bisects the chord. Equation of diameter of parabola y 2  x corresponding to the chord x  y  1  0 is (a) 2y  3 (b) 2y  1 (c) 2y  5 2a 1 Equation of diameter of parabola is y  , Here a  , m  1  y  m 4 [Rajasthan PET 2003] (d) y 1 1 4  2y  1 1 2. E3 Solution: (b) Every diameter of a parabola is parallel to its axis. 60 Note 2a m 5.1.19 Length of Tangent, Subtangent ,Normal and Subnormal. Let the parabola y 2  4 ax. Let the tangent and normal at P(x 1 , y 1 ) meet the axis of parabola at T and G ID respectively, and tangent at P(x 1 , y 1 ) makes angle  with the positive direction of x-axis. A(0, 0 ) is the vertex of the parabola and PN  y. Then, Y yy1=2a(x+x1) U (1) Length of tangent  PT  PN cosec   y 1 cosec  (2) Length of normal  PG  PNcosec (90 o   )  y 1 sec  X T(–x1,0)  S(a,0) A (3) Length of subtangent  TN  PN cot   y 1 cot  D YG (x1,y1) P N G(x1,2a,0) (x1,0) Y (4) Length of subnormal  NG  PN cot(90   )  y 1 tan  X y2=4ax o where , tan   2a  m , [slope of tangent at P(x, y)] y1 Length of tangent, subtangent, normal and subnormal to y2 = 4ax at (at2, 2at) (1) Length of tangent at (at 2 ,2at)  2at cosec   2at (1  cot 2  )  2at 1  t 2 U (2) Length of normal at (at 2 , 2at)  2at sec  2at (1  tan 2  )  2a t 2  t 2 tan 2   2a (t 2  1) (3) Length of subtangent at (at 2 ,2at)  2at cot   2at 2 ST (4) Length of subnormal at (at 2 ,2at)  2at tan   2a Example: 38 The length of the subtangent to the parabola y 2  16 x at the point whose abscissa is 4, is (a) 2 (b) 4 (c) 8 (d) None of these Solution: (c) Since the length of the subtangent at a point to the parabola is twice the abscissa of the point. Therefore, the required length is 8. Example: 39 If P is a point on the parabola y 2  4 ax such that the subtangent and subnormal at P are equal, then the coordinates of P are Solution: (a) (a) (a, 2a) or (a,2 a) (b) (2a, 2 2 a) or (2a,2 2 a) (c) (4 a,  4 a) or (4 a, 4 a) (d) None of these Since the length of the subtangent at a point on the parabola is twice the abscissa of the point and the length of the subnor mal is equal to semi-latus-rectum. Therefore if P(x , y ) is the required point, then 2 x  2a  x  a 168 Conic Section : Parabola Now (x, y) lies on the parabola y 2  4 ax  4 a 2  y 2  y  2 a Thus the required points are (a, 2a) and (a,  2a). 5.1.20 Pole and Polar. The locus of the point of intersection of the tangents to the parabola at the ends of a chord drawn from a fixed point P is called the polar of point P and the point P is called the pole of the polar. contact and is given by yy 1  2a(x  x 1 ) Q T Q E3 T T Polar (h,k) Q (h,k) 60 Equation of polar: Equation of polar of the point ( x 1 , y 1 ) with respect to parabola y 2  4 ax is same as chord of R Pole P(x1,y1) Pole P (x1,y1) R R Polar R Q T ID (1) Polar of the focus is directrix: Since the focus is (a, 0) Equation of polar of y 2  4 ax is y. 0  2a(x  a)  x  a  0, which is the directrix of the parabola y 2  4 ax. U (2) Any tangent is the polar of its point of contact: If the point P(x 1 y1 ) be on the parabola. Its polar and tangent at P are identical. Hence the tangent is the polar of its own point of contact. (x ,y ) 1 Coordinates of pole: The pole of the line lx  my  n  0 with respect to the parabola  n  2am  y  4 ax is  , . l  l D YG 2 1 P Q Q R R  y y y  y2  (i) Pole of the chord joining ( x 1 , y 1 ) and ( x 2 , y 2 ) is  1 2 , 1  which is the same as the point of intersection 2   4a of tangents at ( x 1 , y 1 ) and (x 2 , y 2 ). (ii) The point of intersection of the polar of two points Q and R is the pole of QR. U 5.1.21 Characterstics of Pole and Polar. (1) Conjugate points: 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 ST through P(x 1 , y1 ) and such points are said to be conjugate points. Two points P( x 1 , y1 ) and Q(x 2 , y 2 ) are conjugate points with respect to the parabola y 2  4 ax , if y 1 y 2  2a(x 1  x 2 ). (2) Conjugate lines: If the pole of a line ax  by  c  0 lies on the another line a1 x  b1 y  c1  0 , then the pole of the second line will lie on the first and such lines are said to be conjugate lines. Two lines l1 x  m 1 y  n1  0 and l 2 x  m 2 y  n 2  0 are conjugate lines with respect to parabola y 2  4 ax , if (l1 n 2  l 2 n1 )  2am 1 m 2 Note : The chord of contact and polar of any point on the directrix always passes through focus.  The pole of a focal chord lies on directrix and locus of poles of focal chord is the directrix.  The polars of all points on directrix always pass through a fixed point and this fixed point is focus. Conic Section : Parabola 169 The pole of the line 2 x  y with respect to the parabola y 2  2 x is (a) Solution: (a) 1  (b)  , 0  2   1   0,  2  1   0,   2  (d) None of these Let (x1, y1 ) be the pole of line 2 x  y w.r.t. parabola y 2  2 x its polar is yy 1  x  x1 Also polar is y  2 x ,  Example: 41 (c) y1 1 1 x  1   1 , x1  0, y1 . So Pole is  0,  1 2 0 2  2 60 Example: 40 If the polar of a point with respect to the circle x 2  y 2  r 2 touches the parabola y 2  4 ax , the locus of the pole is [EAMCET 1995] (a) r x a (b) x2  r y a 2 (c) y2  2 r x a (d) x2  r2 y a y   x1 x r2 r2  y  mx  c, where m   1 ; c  x y1 y1 y1 y1 This touches the parabola y 2  4 ax , If c  Required locus of pole (x 1 , y 1 ) is E3 Polar of a point (x1 , y1 ) w.r.t. x 2  y 2  r 2 is xx 1  yy 1  r 2 i.e. yy 1   xx 1  r 2 ay a r2 a    1 m y1  x 1 / y1 x1 ID Solution: (a) y2   2 r2  ay  r2  i.e., y 2  x y x a U 5.1.22 Reflection property of a Parabola. Note D YG The tangent (PT ) and normal (PN ) of the parabola y2 = 4ax at. P are the internal and external bisectors of SPM and BP is parallel to the axis of the parabola and BPN  SPN Y :  When the incident ray is parallel to the axis of the parabola, the reflected ray will always pass through the focus. S A R Y Light ray B Reflected ray X N Light ray Light ray Light ray  The equation of the axis of the parabola is y  2  0 , which is parallel to the x-axis. So, a ray parallel to x-axis is parallel to the axis of the parabola. We know that any ray parallel to the axis of a parabola passes through the focus after reflection. Here (0, 2) is the focus. ST Solution: (a) X T P  A ray of light moving parallel to the x-axis gets reflected from a parabolic mirror whose equation is (y  2)2  4(x  1). After reflection, the ray must pass through the point (a) (0, 2) (b) (2, 0) (c) (0, –2) (d) (–1, 2) U Example: 42  M Tangent ***