IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

joyfrancis

Parametric points of the given hyperbola are (root(24)sec $\theta$ , root(18)tan $\theta$

distance of this point from the given line is

(3root24sec $\theta$ + 2root18tan $\theta$ + 1 ) / root(13)....let this be f( $\theta$ )

so for critical points f'( $\theta$ )=0

so f'( $\theta$ ) = 3root(24/13)tan $\theta$ sec $\theta$ + 2root(18/13)sec^2 $\theta$ = 0

((3root24)sin $\theta$ + 2root18) / (root13(cos^2 $\theta$ )) = 0

sin $\theta$ = -2root18/3root24 = -1/root3

so point is (root(24)sec $\theta$ , root(18)tan $\theta$ ) $\equiv$ (6,-3)

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