Question

If the line $x-y+k=0$ is a normal to $y^2=4ax$ then the value of $k$ is

• A. $4a$
• B. $-a$
• C. $-5a$
• D. $-3a$

Answer 1

The correct option is D $-3a$

$y^2=4ax$

$\therefore 2y\left(\frac{dy}{dx}\right)=4a$

$m=$ slope$=\frac{2a}{y}$

$\therefore$ Slope of normal$=\frac{-y}{2a}$ $(m_1{m}_2=-1)$

Slope of normal equation $=-a/b=1$ $(x-y+k=0)$

$\therefore 1=\frac{-y}{2a}$

$\therefore y=-2a$

$\therefore 4a^2=4ax$

$\therefore x=a$

This equation of normal

$y+2a=x-a$

$\therefore x-y-3a=0$

Thus $k=-3a$.