Answer:
Option B
Explanation:
The given equation of the curve is $y^{2}=4ax$ ...(1)
Differentiating both sides of (1) with respect to x, we get
$2y\frac{dy}{dx}=4a;\Rightarrow \frac{dy}{dx}=\frac{4a}{2y}=
\frac{2a}{y}$ ...(2)
If ψ be the angle which the tangent to the curve at (x,y) makes with the positive direction of x-axis then $\tan\psi=\frac{\text{d}y}{\text{d}x}$ or
$\tan\psi= \frac{2a}{y}$ ...(3), [using (2)]
At $x=a$, then from (1),
$y^{2}=4a$.$ a=4a^{2}\Rightarrow y=±2a$
Hence, we get two points (a, 2a) and (a, -2a) on the curve
At $(a, 2a)$, $x =a$, $y=2a$ and let $\psi=\psi_{1}$.
∴ from (3), $\tan\psi_{1}=\frac{2a}{2a}=1=\tan45°$
$\Rightarrow\psi_{1}=45°$
At (a,−2a), x=a, y=−2a and let $\psi=\psi_{2}$.
∴ from (3), $\tan\psi_{2}=\frac{2a}{-2a}=-1=\tan135°$
or $\psi_{2}=135°$
Hence the required angle between tangents to (1) at $(a,2a)$ and $(a,-2a)= \psi_{2}-\psi_{1}= 135°- 45°=90°$
This shows that the tangent lines to (1) at $(a,2a)$ and $(a,-2a)$ are perpendicular to each other.
