Question

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

• A. 0
• B. $ \frac{\pi }{2} $
• C. $ \frac{\pi }{4} $
• D. $ \frac{\pi }{6} $

Answer 1

Answer: (B)

Any line through origin is $y=m x$.
Since, it is a tangent to $y^{2}=4 a(x-a)$,
it will cut it in two coincident points.
So, roots of $m^{2} x^{2}-4 a x+4 a^{2}$ are equal.
$\therefore $ Product of slope $=-1$ i.e., $b^{2}-4 a c=0$
$\Rightarrow 16 a^{2}-16 a^{2} m^{2}=0$
$\Rightarrow m^{2}=1$ or $m=1,-1$
Hence, required angle is right angle i.e., $\frac{\pi}{2}$.