Chapter 4.2 Tangent and Normal PDF

60 Application of Derivatives 185 4.2.1 Slope of the Tangent and Normal. E3 (1) Slope of the tangent : If tangent is drawn on the curve y  f (x ) at point P( x 1 , y1 ) and this tangent makes an angle  with positive x-direction then, y ID  dy   tan  = slope of the tangent    dx ( x1 , y1 )...

60 Application of Derivatives 185 4.2.1 Slope of the Tangent and Normal. E3 (1) Slope of the tangent : If tangent is drawn on the curve y  f (x ) at point P( x 1 , y1 ) and this tangent makes an angle  with positive x-direction then, y ID  dy   tan  = slope of the tangent    dx ( x1 , y1 ) Note :  If tangent is parallel to x-axis   0   dy   dx ( x1 , y1 ) Tangen t Norm al 0  U  dy   If tangent is perpendicular to x-axis       2  dx ( x1 , y1 )  x O D YG (2) Slope of the normal : The normal to a curve at P( x 1 , y1 ) is a line perpendicular to the tangent at P and passing through P and slope of the normal = 1 Slope of tangent =  dx  1    dy    dy  P ( x1 , y1 )    dx  P ( x1 , y1 ) U Note :  If normal is parallel to x-axis ST  dx   dx   0 or   0      dy ( x1 , y1 )  dy ( x 1 , y1 )  If normal is perpendicular to x-axis (for parallel to y-axis)  dy     0  dx ( x 1 , y1 ) Example: 1 The slope of the tangent to the curve x 2  y 2  2c 2 at point (c, c) is (a) 1 Solution: (b) (b) – 1 Given x 2  y 2  2c 2 Differentiating w.r.t. x, 2 x  2 y dy 0 dx (c) 0 [AMU 1998] (d) 2 186 Application of Derivatives  2y The line x  y  2 is tangent to the curve x 2  3  2 y at its point (a) (1, 1) Solution: (a) (b) (–1, 1) (c) ( 3 , 0) Given curve x 2  3  2 y diff. w.r.t. x, 2 x   dy  x dx 2 dy ; dx   x  1 ; x  1 dx  y  1 point (1, 1) The tangent to the curve y  2 x 2  x  1 at a point P is parallel to y  3 x  4 , the co-ordinate of P are [Rajasthan P (a) (2, 1) Solution: (b) (b) (1, 2) Given y  2 x  x  1 2 ID Example: 3 (d) (3, – 3) E3 Slope of the line = – 1 dy [MP PET 1998] 60 Example: 2 dy dy  x  dy   2 x      1  dx dx y  dx (c, c) (c) (– 1, 2) (d) (2, – 1) Clearly 4 h  1  3 D YG h  1  k  2. P is (1, 2). U  dy  Let the co-ordinate of P is (h, k) then    4h  1  dx (h, k ) 4.2.2 Equation of the Tangent and Normal. (1) Equation of the tangent : We know that the equation of a line passing through a point and having slope m is y  y 1  m (x  x 1 ) P(x1 , y1 ) U  dy  Slope of the tangent at ( x 1 , y 1 ) is =    dx  ( x 1 , y1 ) ST The equation of the tangent to the curve y  f (x ) at point P(x 1 , y 1 ) is  dy  y  y1    (x  x 1 )  dx  ( x1 , y1 ) (2) Equation of the normal : Slope of the Normal = 1 dy      dx  ( x 1 , y1 ) Thus equation of the normal to the curve y  f (x ) at point P(x 1 , y 1 ) y  y1  1  dy     dx  ( x 1 , y1 ) (x  x 1 ) Application of Derivatives 187 Note :  If at any point P(x 1 , y 1 ) on the curve y  f (x ) , the tangent makes equal angle with the axes, then at the point P,   3 dy. Hence, at P tan    1. 4 dx The equation of the tangent at (4,  4) on the curve x 2  4 y is (b) 2 x  y  12  0 (a) 2 x  y  4  0 Solution: (d) 4 or x 2  4 y  2 x  4 [Karnataka CET 2001] (d) 2 x  y  4  0 (c) 2 x  y  4  0 dy x dy  dy       2. dx 2 dx  dx  (4 , 4 ) 60 Example: 4   dy  We know that equation of tangent is (y  y 1 )    (x  x 1 )  y  4  2(x  4 )  2 x  y  4  0.  dx  ( x1 , y1 ) y  sin x 2  1 0 (d) y  1  2  ( x  1) ( x  1)  x  1. U The equation of the tangent to the curve y  be  x / a at the point where it crosses y-axis is (b) ax  by  1 (a) ax  by  1 Curve is y  be  x / a (c) x y  1 a b D YG Solution: (d) (c) y  x dy    dy   cos x    0 dx 2 2  dx  (1, 1)  Equation of normal is y  1  Example: 6 at (1, 1) is 2 (b) x  1 (a) y  1 Solution: (b) x E3 The equation of the normal to the curve y  sin ID Example: 5 (d) x y  1 a b Since the curve crosses y-axis (i.e., x  0 )  y  b Now dy b  x / a b  dy   e. At point (0, b),    dx a a  dx (0, b )  Equation of tangent is y  b  Example: 7 If the normal to the curve y  f (x ) at the point (3, 4) makes an angle U f (3 ) is equal to ST Slope of the normal  3 with the positive x-axis then 4 [IIT Screening 2000; DCE 2001] (a) – 1 Solution: (d) x y b  1. (x  0)  a b a (b) 3 4 (c) 4 3 (d) 1 3 1 1  tan  4  dy  dy / dx    dx (3, 4 )  dy   1 ; f (3)  1.     dx (3, 4 ) Example: 8 The point (s) on the curve y 3  3 x 2  12 y where the tangent is vertical (parallel to y-axis), is are[IIT Screenin  4  ,  2 (a)  3   Solution: (d) y 3  3 x 2  12 y  11  ,1 (b)     3   (c) (0, 0)  4  , 2  (d)   3   188 Application of Derivatives  3y 2. dy dy dy dy 6x dx 12  3 y 2 (3 y 2  12 )  6 x  0   6 x  12.     dx dx dx dy 6x dx 12  3 y 2 Tangent is parallel to y-axis, 4 dx , for y  2  0  12  3 y 2 = 0 or y  2. Then x   dy 3 At which point the line x y   1 touches the curve y  be  x / a a b (a) (0, 0) Solution: (c) (b) (0, a) Let the point be (x 1 , y1 )  y1  be Also, curve y  be  x / a  (c) (0, b) x1 / a......(i) dy b  x / a  e dx a b  x 1 / a y1  dy   e    a a  dx ( x 1 , y1 ) ID U The abscissa of the point, where the tangent to curve y  x 3  3 x 2  9 x  5 is parallel to x-axis are [Karnataka C (b) x  1 and 1 (c) x  1 and 3 (d) x  1 and 3 D YG (a) 0 and 0 Solution: (d) y1 x x y (x  x 1 )    1 1 a a y1 a x x y   1 , we get, y1  b and 1  1  1  x 1  0 a b a Hence, the point is (0, b). Example: 10 (d) (b, 0) (by (i)) Now, the equation of tangent of given curve at point (x 1 , y1 ) is y  y1  Comparing with [Rajasthan PET 1999] E3 Example: 9 60  4  y  2 does not satisfy the equation of the curve,  The point is   , 2  3   dy  3x 2  6x  9. y  x 3  3x 2  9x  5  dx We know that this equation gives the slope of the tangent to the curve. The tangent is parallel to xdy 0 axis dx Therefore, 3 x 2  6 x  9  0  x  1, 3. 4.2.3 Angle of Intersection of Two Curves. U The angle of intersection of two curves is defined to be the angle between the tangents to the two curves at their point of intersection. ST We know that the angle between two straight lines having slopes m 1 yand m 2 y = f2x y = f1x m  m2   tan 1 1 1  m 1m 2 P Also slope of the tangent at P(x 1 , y1 )  dy  m1    ,  dx 1( x1 , y1 )  dy  m2     dx  2( x1 , y1 )  O Thus the angle between the tangents of the two curves y  f1 (x ) and y  f2 (x ) x Application of Derivatives 189  dy   dy      dx  1( x1 , y1 )  dx  2( x1 , y1 ) tan    dy   dy  1    dx  1( x1 , y1 )  dx  2 ( x1 , y1 ) orthogonal, then   60 Orthogonal curves : If the angle of intersection of two curves is right angle, the two curves are said to intersect orthogonally. The curves are called orthogonal curves. If the curves are  2 The angle between the curves y 2  x and x 2  y at (1, 1) is (a) tan 1 (b) tan 1 3 4 (c) 90 o Given curve y 2  x and x 2  y Differentiating w.r.t. x, 2 y (d) 45 o ID Solution: (b) 4 3 dy dy  1 and 2 x  dx dx 1  dy   dy  and    2    dx (1, 1)  dx (1, 1) 2 U Example: 11 E3  dy   dy  m 1m 2  1       1  dx 1  dx  2 Example: 12 If the two curves y  a x and y  b x intersect at  , then tan  equal (a) Solution: (a) D YG Angle between the curve 1 2 2  tan   3    tan 1 3.  tan   1 4 4 1 .2 2 log a  log b 1  log a log b (b) log a  log b 1  log a log b U Slope of tangent of second curve, m 2  (d) None of these dy  dy   a x log a     m 1  log a dx  dx (0 , 1) dy  dy   b x log b  m 2     log b dx  dx (0, 1) m1  m 2 log a  log b . 1  m 1m 2 1  log a log b ST  tan   The angle of intersection between curve xy  6 and x 2 y  12 3 (a) tan 1   4 Solution: (b) log a  log b 1  log a log b Clearly the point of intersection of curves is (0, 1) Now, slope of tangent of first curve, m 1  Example: 13 (c) [MP PET 2001]  3  (b) tan 1    11   11  (c) tan 1    3  The equation of two curves are xy  6 and x 2 y  12 from (i) we obtain y  (d) 0 o 6 putting this value of y in x 6 equation (ii) to obtain x 2    12  6 x  12  x  2 x Putting x  2 in (i) or (ii) we get, y  3. Thus, the two curves intersect at P(2, 3) Differentiating (i) w.r.t. x, we get x dy dy y 3  dy  y 0      m1    dx dx x 2  dx (2, 3 ) 190 Application of Derivatives Differentiating (ii) w.r.t. x, we get x 2 dy 2 y dy   2 xy  0  dx dx x m1  m 2  dy       3  m 2  tan    dx (2, 3 ) 1  m1m 2 3   3  3 3    tan 1.  3   1    (3)   11  2    2   11 4.2.4 Length of Tangent, Normal, Subtangent and Subnormal. 60 Let the tangent and normal at point P( x , y ) on the curve y  f (x ) meet the x-axis at points A 2 y ID  dy  1   dx  (1) Length of tangent PA  ycosec   y  dy     dx  E3 and B respectively. Then PA and PB are called length of tangent and normal respectively at point P. If PC be the perpendicular from P on x-axis, the AC and BC are called length of subtangent and dy subnormal respectively at P. If PA makes angle  with x-axis, then tan   from fig., we find dx that 2 U  dy  (2) Length of normal PB  y sec   y 1     dx  D YG y (3) Length of subtangent AC  y cot    dy     dx  Tangen t Norm al P (x, y)   O A C B x  dy  (4) Length of subnormal BC  y tan   y    dx  Example: 14 The length of subtangent to the curve x 2 y 2  a 4 at the point ( a, a) is (a) 3 a (c) a (d) 4a Equation of the curve x y  a. U Solution: (c) (b) 2a [Karnataka CET 2001] 2 2 4 Differentiating the given equation, dy dy y  dy   a   y 2 2x  0         1 dx x dx dx   (  a, a ) a ST x 2 2y Therefore, sub-tangent = Example: 15 For the curve y n  an 1 x , the sub-normal at any point is constant, the value of n must be (a) 2 Solution: (a) y a.  dy     dx  y n  an 1 x  ny n 1 (b) 3 dy  dy  an 1  a n  1     n 1 dx  dx  ny (c) 0 (d) 1 Application of Derivatives 191  Length of the subnormal = y dy ya n 1 an 1 y 2 n   dx ny n 1 n We also know that if the subnormal is constant, then a n 1 2  n.y should not contain y. n 60 Therefore, 2  n  0 or n  2. 4.2.5 Length of Intercept made on Axis by the Tangent.  dy  Equation of tangent at any point ( x 1 , y1 ) to the curve y  f (x ) is y  y1    (x  x1 )  dx ( x1 , y1 ) E3 Equation of x-axis y = 0......(ii) and Equation of y-axis x = 0......(iii)     y1   Solving (i) and (ii) we get x  x 1    dy        dx ( x 1 , y1 )  ID Y R D YG U     y1   x-intercept OQ  x 1    dy        dx ( x 1 , y1 ) ......(i) P (x1 y1) O Q X   dy   Similarly solving (i) and (iii) we get, y-intercept OR = y1   x1      dx ( x 1 , y1 )  Example: 16 The sum of intercepts on co-ordinate axes made by tangent to the curve (a) a Solution: (a) x  (b) 2a 1 y  a   U 2 x (c) 2 a y dy dy  0   dx x 2 y dx y ( X  x ) or X y  Y x  xy ( x  y )  axy or x ST (d) None of these 1 Hence tangent at (x, y) is Y  y   Clearly its intercepts on the axes are Sum of the intercepts = x  y  a is X a x a x and  Y 1. a y a y. a( x  y )  a. a  a. 4.2.6 Length of Perpendicular from Origin to the Tangent. Length of perpendicular from origin (0, 0) to the tangent drawn at point P(x 1 , y 1 ) of the curve y  f (x ) p  dy  y1  x 1    dx  ( x 1 , y1 )  dy  1   dx  2 192 Application of Derivatives Example: 17 is The length of perpendicular from (0, 0) to the tangent drawn to the curve y 2  4 (x  2) at point (2, 4) (a) 1 2 Solution: (c) (b) 3 (c) 5 6 (d) 1 5 Differentiating the given equation w.r.t. x , 2 y dy dy 1   4 at point (2, 4) dx dx 2 ST U D YG U ID E3 60 1  dy  4  2  y1  x 1   6 2  dx  = . P 2 1 5 dy   1 1  4  dx  60 Application of Derivatives 193 Basic Level If the line y  2 x  k is a tangent to the curve x 2  4 y, then k is equal to 2. (b) 1 1 (b)  ,  4 2 (d) x  y  3  0 dy 1 dx dx 0 dy (c) [Rajasthan PET 2000] (d) None of these (b) x  5 y  2 U   (b) y  2  2  x   4  (c) 5 x  y  2 (d) 5 x  y  2  0  is 4   (c) y  2   2  x   4    (d) y  2  2  x   4  ST For the curve x  t 2  1, y  t 2  t , the tangent line is perpendicular to x-axis where (b) t   (c) t  1 3 [MNR 1980] (d) t   1 3 If at any point on a curve the sub-tangent and subnormal are equal, then the tangent is equal to (a) Ordinate 9. (b) The equation of tangent to the curve y  2 cos x at x  (a) t  0 8. (d) (1, 1) The equation of the tangent to the curve (1  x 2 )y  2  x , where it crosses the x-axis, is   (a) y  2  2 2  x   4  7. (c) x  y  1  0 (b) x  y  1  0 D YG dy 0 dx (a) x  5 y  2 6. 1 2 [Rajasthan PET 1990, 92] If normal to the curve y  f (x ) is parallel to x-axis, then correct statement is (a) 5. (c) (4, 2) [AMU 2002] If x  t 2 and y  2 t , then equation of the normal at t  1 is (a) x  y  3  0 4. (d)  (c) – 4 The point on the curve y 2  x where tangent makes 45 o angle with x-axis is 1 1 (a)  ,  2 4 3. 1 2 ID (a) 4 U 1. E3 Tangent and Normal (b) 2 ordinate (c) 2 (ordinate) (d) None of these If the tangent to the curve 2 y 3  ax 2  x 3 at the point (a, a) cuts off intercepts,  and  on the coordinate axes such that  2   2  61 , then a = (a) 30 10. (b)  5 (c)  6 If the tangent to the curve x  a(  sin  ), y  a(1  cos  ) at    3 (d)  61 makes an angle  with x-axis, then  = 194 Application of Derivatives (a) (b) (c)  6 The fixed point P on the curve y  x 2  4 x  5 such that the tangent at P is perpendicular to the line x  2 y  7  0 (a) (3, 2) (b) (1, 2) ID (c) x   2 (d) 1 t  y  y1 1 (b) x  x1 a y  a y1 D YG x (d) No where x  y  a at the point (x1 , y1 ) is (c) x x1  y y1  a (b) (1, 4) (c) (0, 0) (d) None of these (b) 6 3 (d) (– 4, 4) x  y  4 at point (4, 4) on coordinate axes is The sum of the intercepts made by a tangent to the curve (c) 8 2 (d) 256 The angle of intersection between the curve y 2  16 x and 2 x 2  y 2  4 is U (b) 30 o ST The equation of normal to the curve ax by   a2  b 2 sec  tan  (b) (c) 45 o [Rajasthan PET 1993] (d) 90 o x2 y2   1 at the point ( a sec  , b tan  ) is a2 b 2 ax by   a2  b 2 sec  tan  (c) ax by   a2  b 2 sec  tan  (d) ax by   ab sec  tan  (d) dy  1 dx If tangent to a curve at a point is perpendicular to x-axis, then at the point (a) dy 0 dx (b) dx 0 dy (c) dy 1 dx If m be the slope of a tangent to the curve e y  1  x 2 then (a) | m |  1 23. [Rajasthan PET 1993] A tangent to the curve y  x 2  3 x passes through a point (0, – 9) if it is drawn at the point (a) 22.  The equation of tangent to the curve (a) 0 o 21. (c)  t (b) x  4 (a) 4 2 20. 1 t U  (a) (– 3, 0) 19. E3 (b) x1 18. (d) None of these The slope of the curve y  sin x  cos 2 x is zero at the point, where (a) 17. (d) None of these The slope of the tangent to the curve y 2  4 ax drawn at point (at 2 , 2 at) is (a) x  16. (c) x 2  y 2  x 2 y 2 (b) x 2  y 2  x 2 y 2 (a) t 15. (c) (2, 1) The points of contact of the tangents drawn from the origin to the curve y  sin x lie on the curve (a) x 2  y 2  xy 14. 5 6 (d) a  1, b  2 (c) a  1, b  2 (b) a  1, b  2 is given by 13. (d) If the tangent to the curve xy  ax  by  0 at (1, 1) is inclined at an angle tan 1 2 with x-axis, then (a) a  1, b  2 12. 2 3 60 11.  3 (b) m  1 (c) | m |  1 (d) | m |  1 The equation of the tangent to the curve y  e | x| at the point where the curve cuts the line x  1 is (a) x  y  e (b) e (x  y)  1 (c) y  ex  1 (d) None of these Application of Derivatives 195 The slope of the tangent to the curve y   1 x x 0 (a) (b) 1 (b) x  2 y  2  0  2 4  3 1 (b) tan  4 2 tan 1   7  (d) None of these  1  (c)   , 0   2  (d) Non where U (b) x sec t  y co sec t  a (c) x cosec t  y sec t  a [Rajasthan PET 1988] (d) xco sect  y sec t  a D YG t  The length of the tangent to the curve x  a  cos t  log tan  , y  a sin t is 2   (b) ay (c) a The point at the curve y  12 x  x 3 where the slope of the tangent is zero will be (a) (0, 0) (b) (2, 16) (c) (3, 9) The angle of intersection between the curves y  x 2 and 4 y  7  3 x 3 at point (1, 1) is (b) U  4  3 (c)  2 (d) xy [Rajasthan PET 1992] (d) None of these [Andhra CEE 1992] (d) None of these Advance Level ST (a) 33.     The equation of the tangent to the curve x  a cos 3 t, y  a sin 3 t a t ' t' point is (a) ax 32. (d) 2 x  y  4  0 ID (b) (2, 0) (a) x sec t  y cosec t  a 31. [Rajasthan PET 1989, 1992] Tangent to the curve y  e 2 x at point (0, 1) meets x-axis at the point (a) (0, a) 30. (c) [Rajasthan PET 1997] (d) 90 o (c) 2 x  y  4 The angle of intersection of the curve y  4  x 2 and y  x 2 is (a) 29. (d) None of these The equation of the normal to the curve y  x (2  x ) at the point (2, 0) is [Rajasthan PET 1989, 1993; MNR 1978] 28. 1 4 (c) 60 o (b) 45 o (a) x  2 y  2 27. (c) The angle of intersection between the curves x 2  4 ay and y 2  4 ax at origin is (a) 30 o 26. at the point where x  1 is 3 60 25. 1 2 dx E3 24. Consider the following statements: Assertion (A) : The circle x 2  y 2  1 has exactly two tangents parallel to the x-axis Reason (R) : dy  0 on the circle exactly at the points (0,  1). Of these statements dx (a) Both A and R and true and R is the correct explanation of A (b) Both A and R are true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true 34. The slope of the tangent to the curve x  3 t 2  1, y  t 3  1 at x  1 is [SCRA 1996] 196 Application of Derivatives (a) 0 At what points of the curve y  1 4  (b)  ,  and (-1, 0) 2 9 (b) x  1  1 D YG 2 (c) x  2  3 1 (d) x  1 3 The length of the normal at point 't' of the curve x  a(t  sin t), y  a(1  cos t) is t t (b) 2 a sin 3   sec   2 2 U (b) a 2 [Rajasthan PET 2001] t t (c) 2 a sin  tan   2 2 The length of normal to the curve x  a(  sin ), y  a(1  cos  ) at the point   (c) t (d) 2 a sin   2  is 2 2a [Rajasthan PET 1999; (d) a 2 ST The area of the triangle formed by the coordinate axes and a tangent to the curve xy  a 2 at the point (x 1 , y1 ) on it is (a) a2 x1 y1 [DCE 2001] (b) a 2 y1 x1 (c) 2a 2 (d) 4 a 2 The normal of the curve x  a(cos    sin ), y  a(sin   cos  ) at any  is such that [DCE 2000] (a) It makes a constant angle with x-axis (b) It passes through the origin (c) It is at a constant distance from the origin (d) None of these An equation of the tangent to the curve y  x 4 from the point (2, 0)not on the curve is (a) y  0 46. (d) (e 2 ,  2 e 2 ) (c) (e 2 , 2 e 2 ) The abscissa of the points of curve y  x (x  2)(x  4 ) where tangents are parallel to x-axis is obtained as (a) 2a 45. ID (b) (e, e) AIEEE 2004] 44. 3  (d)  , 2  2  (c) (5, – 2) Coordinates of a point on the curve y  x log x at which the normal is parallel to the line 2 x  2 y  3 are [Rajasthan PET 2 (a) asin t 43. 1  1 1 (d)  ,  and   3,  2  3 7 (d) y 3 (c) y 2  1  (b)   ,  2   2  3 42. 1  1 1  (c)  ,  and   1,  3   3 47  [UPSEAT 1999] The point of the curve y 2  2(x  3) at which the normal is parallel to the line y  2 x  1  0 is (a) x  2  41. (d) None of these 2 3 1 2 x  x , tangent makes the equal angle with axis 3 2 (b) x 3 (a) (0, 0) 40. (c) 6 For the curve xy  c 2 the subnormal at any point varies as (a) (5, 2) 39. 6 7 [MNR 1994] 60 (b) (a) x 2 38. (d) 2 E3 22 7 1  1 5  (a)  ,  and   1,   6   2 24  37. (c)  The slope of tangent to the curve x  t 2  3 t  8, y  2 t 2  2 t  5 at the point (2,  1) is (a) 36. 1 2 U 35. (b) (b) x  0 (c) x  y  0 For the curve by 3 (x  a)3 the square of subtangent is proportional to (d) None of these Application of Derivatives 197 (a) (Subnormal)1/2 47. (d) None of these The tangent to the curve y  ax 2  bx at (2,  8) is parallel to x-axis. Then (a) a  2, b  2 48. (c) (Subnormal)3/2 (b) Subnormal [AMU 1999] (d) a  4, b  4 (c) a  2, b  8 (b) a  2, b  4 If the area of the triangle include between the axes and any tangent to the curve x n y  a n is constant, then n is (a) 1 (b) a 2  b 2  l 2  m 2 (d) ordinate D YG (b) (1, 2) 1 1 (c)  ,   2 2 1  1 (d)  ,   8 16     at any point varies as the   (c) Square of the abscissa of the point (d) Square of the ordinate of the point If the parametric equation of a curve given by x  e t cos t, y  e t sin t, then the tangent to the curve at the point U  4 makes with axes of x the angle ST (b)  4 (c)  3 (d)  2 For the parabola y 2  4 ax , the ratio of the subtangent to the abscissa is (b) 2 : 1 (c) x : y (d) x 2 : y Tangents are drawn from the origin to the curve y  cos x. Their points of contact lie on (a) x 2 y 2  y 2  x 2 (b) x 2 y 2  x 2  y 2 (c) x 2 y 2  x 2  y 2 (d) None of these If y  4 x  5 is a tangent to the curve y 2  px 3  q at (2, 3) then (a) p  2, q  7 58. (c) abscissa (b) Ordinate of the point (a) 1 : 1 57. (d) a 2  b 2  l 2  m 2 (a) Abscissa of the point (a) 0 56. (c) a 2  b 2  l 2  m 2 U (b) (Ordinate)2  e x / a  e x / a The length of the normal to the curve y  a 2  t 55. (d) None of these The point P of the curve y 2  2x 3 such that the tangent at P is perpendicular to the line 4 x  3 y  2  0 is given (a) (2, 4) 54. (c) Line x The length of the normal at any point on the catenary y  c cos h  varies as c by 53. 1 2 x2 y2 x2 y2  2  1 and 2  2  1 cut each other orthogonally, then 2 a l m b (a) (abscissa)2 52. (d) E3 If the curves (b) Parabola (a) a 2  b 2  l 2  m 2 51. 3 2 x  All points on the curve y 2  4 a  x  a sin  at which the tangents are parallel to the axis of x, lie on a a  (a) Circle 50. (c) ID 49. (b) 2 60 equal to (b) p  2, q  7 (c) p  2, q  7 (d) p  2, q  7 The curve y  e xy  x  0 has a vertical tangent at the point (a) (1, 1) (b) At no point (c) (0, 1) (d) (1, 0) 198 Application of Derivatives If the tangent and normal at any point P of parabola meet the axes at T and G respectively then (a) ST = SG.SP 60. Slope of the tangent to the curve y | x 3 | at origin is (a)  2 (b)  3 (c) n (d) (2, –2 ) x2 y2   1 at the point (8, 3 3 ) is 16 9 (b) x  y  25 (c) y  2 x  25 [MP PET 1996] (d) 2 x  3 y  25 D YG (b) 30 o (c) 45 o [Rajasthan PET 1998] (d) 90 o The subtangent to the curve x m y n  a m n at any point is proportional to (b) Abscissa (c) (Ordinate)n [Rajasthan PET 1998] (d) (Abscissa)n If tangents drawn on the curve x  at 2 , y  2at is perpendicular to x-axis then its point of contact is (b) (a, 0) (c) (0, a) (d) (0, 0) Tangents are drawn to the curve y  x 2  3 x  2 at the points where it meets x-axis. Equations of these tangents U are (a) x  y  2  0, x  y  1  0 (b) x  y  1  0, x  y  2 [Rajasthan PET 1993] (c) x  y  1  0, x  y  0 (d) x  y  0, x  y  0 If the tangents at any point on the curve x 4  y 4  a 4 cuts off intercept p and q on the axes, the value of ST 69. (d) 2a 2 The angle of intersection between the curves xy  a 2 and x 2  y 2  2a 2 is (a) (a, a) 68. (c) (–1, 6) ID 3 x  2 y  25 (a) Ordinate 67. (b) (1, 0) The equation of normal to the curve (a) 0 o 66. (c) a 2 (b) 2a The point of the curve y  x 2  3 x  2 at which the tangent is perpendicular to the y  x will be (a) 65. (d) For non-zero values of The sum of the squares of intercepts made by a tangent to the curve x 2 / 3  y 2 / 3  a 2 / 3 with coordinate axes is [Rajasthan (a) (0, 2) 64. (c) 3 [Rajasthan PET 1998] E3 (b) 2 (a) a 63. n (d) 0 x y x y The line       2 , touches the curve       2 at point (a, b) then n   a b a b (a) 1 n 62.  6 U 61. (d) ST = SG  SP (c) ST  SG = SP (b) ST = SG = SP 60 59. p 4 / 3  q 4 / 3 is (a) a 4 / 3 70. (b) a 1 / 2 (c) a 1 / 2 (d) None of these At any point (x 1 , y 1 ) of the curve y  ce x / a (a) Subtangent is constant (b) Subnormal is proportional to the square of the ordinate of the point (c) Tangent cuts x-axis at (x 1  a) distance from the origin (d) All the above 71. The equation of the tangent to the curve y  1  e x / 2 at the point where it meets y-axis is Application of Derivatives 199 (a) x  2 y  2 72. (c) x  y  2 (b) 2 x  y  0 (d) None of these The coordinates of the points on the curve x  a(  sin  ), y  a(1  cos  ) , where tangent is inclined an angle  to 4 the x-axis are For a curve (Length of normla )2 (Length of tangent ) 2 is equal to (c) 2, 1 (d) –2, 1 Let C be the curve y 3  3 xy  2  0. If H and V be the set of points on the curve C where tangent to the curve is U (a) H  {(1, 1)}, V   (b) H  , V  {(1, 1)} (c) H  {(0, 0)}, V  {(1,1)} (d) None of these D YG If the line ax  by  c  0 is a normal to the curve xy  1 then (a) a, b  R 78. 2 9 (Subtangent)/(Subnormal) (c) ID (b) –1, 1 horizontal and vertical respectively, then 77. (d) If the curve y  x 2  bx  c , touches the line y  x at the point (1, 1), the values of b and c are (a) – 1, 2 76. 2 9 (c)  (b) 1 (a) (Subnormal)/(Subtangent) (b) (Subtangent/Subnormal)2 (d) Constant 75. 60 If equation of normal at a point (m 2  m 3 ) on the curve x 3  y 2  0 is y  3mx  4 m 3 , then m 2 equals (a) 0 74.    (d)  a, a   1   2    E3 73.     (c)  a   1  , a  2         (b)  a   1  , a  2     (a) (a, a) (b) a  0, b  0 (c) a  0, b  0 or a  0, b  0 (d) a  0, b  0 If the tangent to the curve f (x )  x 2 at any pint (c, f (c)) is parallel to line joining the points (a, f (a)) and (b, f (b)) on the curve, then a, c, b are in (a) H.P. 79. U (b) 2a 2 (c) a2 2 (d) 3a 2 2 ST (b)    (c)  2    (d)  2    The distance between the origin and the normal to the cure y  e 2 x  x 2 at the point x  0 is (a) 2 5 82. (d) A.P. and G.P. both The angle of intersection between the curves r  a sin(   ) and r  b cos(   ) is (a)    81. (c) A.P. The area of triangle formed by tangent to the hyperbola 2 xy  a 2 and coordinates axes is (a) a 2 80. (b) G.P. (b) 2 (c) 5 (d) None of these 5 If the curve y  ax 2  6 x  b passes through (0, 2) and has its tangent parallel to x-axis at x  3 , then the value 2 of a and b are [SCRA 1999] (a) 2, 2 (b) –2, –2 (c) –2, 2 (d) 2, –2 200 Application of Derivatives 83. If at any point S of the curve by 2  (x  a)3 the relation between subnormal SN and subtangent ST be p (SN )  q (ST ) then p/q is equal to 2 [Rajasthan PET 1999; EAMCET 1991] (a) (b) (b) x  3 2 y  104 E3 (d) None of these (c) HP (d) None of these At what point the slope of the tangent to the curve x 2  y 2  2 x  3  0 is zero (b) (3, 0); (1, 2) (c) (–1, 0); (1, 2) [Rajasthan PET 1989, 1995] (d) (1, 2); (1, –2) Let the equation of a curve be x  a (  sin ), y  a(1  cos  ). If  changes at a constant rate k then the rate of 2k (b) k 3 3 is (c) k D YG 3  U (a) (d) None of these The equation of a curve is y  f (x ). The tangents at (1, f (1)), (2, f (2)) and (3, f (3)) makes angles respectively with the positive direction of the x-axis. Then the value of (a)  1 3 (b) 1  3 2 (c) 0 f (x ) f (x )dx   f (x)dx   , 6 3 and  4 3 is equal to 1 (d) None of these 3 1  P(2,2) and Q  ,  1  are two points on the parabolas y 2  2 x. the coordinates of the point R on the parabola, 2  U 90. (d) None of these (c) 3 2 x  y  104 change of the slope of the tangent to the curve at   89. 60 (c) (0, 0) (b) GP (a) (3 0); (–1, 0) 88. (b) (1, 1/3) (1, –1/3) At any point (except vertex) of the parabola y 2  4 ax subtangent, ordinate and subnormal are in (a) AP 87. (d) None of these The equation of the normal to the curve y 2  x 3 at the point whose abscissa is 8, will be (a) x  2 y  104 86. b a (c) The points on the curve 9 y 2  x 3 where the normal to the curve cuts equal intercepts from the axes are (a) (4, 8/3), (4, –8/3) 85. 8a 27 ID 84. 8b 27 where the tangent to the curve is parallel to the chord PQ, is ST 5 5   (a)  , 4  2   91. (d) None of these The number of tangents to the curve x 3 / 2  y 3 / 2  a 3 / 2 ,. where the tangents are equally inclined to the axes, is (a) 2 92. 1 1 (c)  ,  8 2 (b) (2, – 1) (b) 1 (c) 0 (d) 4 If at each point of the curve y  x 3  ax 2  x  1 the tangent is inclined at an acute angle with the positive direction of the x-axis then (a) a  0 (b) a  3 (c)  3  a  3 *** (d) None of these E3 Assignment (Basic and Advance Level) 60 200 Application of Derivatives 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 c b a c a c a b a d b a c b b b a d d a 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 b d d a d a c c b 41 42 43 44 45 46 47 48 c c c c a b c a 61 62 63 64 65 66 67 68 d c b d a b d b 81 82 83 b a a ID 1 55 69 70 71 72 73 74 75 76 77 78 79 80 a d d c d a b b c c a d b d a a b a d c d a 49 50 51 52 53 54 56 57 58 59 60 b c b d d d b c a d b d D YG U c 85 86 87 88 89 90 91 92 a b b d d a c b c ST U 84