A normal to parabola, whose inclination is `30^(@)`, cuts it again at an angle of (a) `tan^(-1)((sqrt(3))/(2))` (b) `tan^(-1)((2)/(sqrt(3)))` (c) `tan^(-1)(2sqrt(3))` (d) `tan^(-1)((1)/(2sqrt(3)))`

Options

A. `tan^(-1)((sqrt(3))/(2))`
B. `tan^(-1)((2)/(sqrt(3)))`
C. `tan^(-1)(2sqrt(3))`
D. `tan^(-1)((1)/(2sqrt(3)))`

Answer

D

Explanation

The normal at `P (a t_(1)^(2), 2a t_(1))` is `y + xt_(1) =2a t_(1) + a t_(1)^(3)` with slope say `tan alpha = - t_(1) = (1)/(sqrt(3))`.
If it meets curve at `Q (a t_(2)^(2), 2a t_(2))` then `t_(2) =- t_(1) -(2)/(t_(1)) =- (7)/(sqrt(3))`.
Then angle `theta` between parabola (tangent at Q) and normal at P is given by
`tan theta = |(-t_(1)-(1)/(t_(2)))/(1-(t_(1))/(t_(2)))| =(1)/(2sqrt(3))`
`rArr theta = tan^(-1) ((1)/(2sqrt(3)))`