Question

The line $y=\sqrt{2}x+4\sqrt{2}$ is a normal to $y^2=4ax$ then a =

• A. $2$
• B. $\sqrt{2}$
• C. $1$
• D. $-1$

Answer 1

The correct option is D $-1$

The general form of equation of normal to the parabola $y^2=4ax$ is $y=mx-2am-a{m}^3$.....(1) where $m$ is the slope of the normal.

According to the problem, $y=\sqrt{2}x+4\sqrt{2}$.....(2) is the normal to the parabola.

Equation (1) and (2) are identical.

Then $m=\sqrt{2}$ and $-2am-a{m}^3=4\sqrt{2}$ or, $a(-2\sqrt{2}-2\sqrt{2})=4\sqrt{2}$ or, $a=-1$ [ Using the value of $m$].

So $a=-1$.