The angle between the tangents draw from the origin to the parabola $y^2=4a(x-a)$ is

Question

$\begin{array}{1 1}(A)\;90^{\circ} \\(B)\;30^{\circ} \\(C)\;\tan^{-1}(1/2) \\(D)\;45^{\circ} \end{array}$

Answer 1

Any line through origin is $y=mx$ .

Since it is a tangent to $y^2=4a(x-a)$

it will cut it in two coincident points.

Roots of $m^2x^2-4ax+4a^2$ are $B^2-4AC=0$

$16a^2-16a^2m^2=0$

(or) $m^2=1 \;(or)\; m=1,-1$

Hence it is right angle.

Hence A is the correct answer.