Question

Normal to the parabola $y^2=4ax$ where $m$ is the slope of the normal is

• A. $y=mx+2am-a{m}^3$
• B. $y=mx-2am-a{m}^3$
• C. $y=mx-2am+a{m}^3$
• D. none of these

Answer 1

Answer: (B)

$y=mx-2am-a{m}^3$

Given, $y^2=4ax$

$\Rightarrow 2ydy=4a$
$\Rightarrow \frac{dy}{dx}=\frac{2a}{y}$
Thus slope of normal to the given parabola at any point is,
$=-\left(\frac{1}{\frac{dy}{dx}}\right)=-\frac{y}{2a}=m$ (given)
$\Rightarrow y=-2am$
Thus from $y^2=4ax,$ $x=\frac{y^2}{4a}=a{m}^2$
Therefore the point is $(a{m}^2,-2am)$
Hence the required line passing through $(a{m}^2,-2am)$ and having slope $m$ is,
$(y+2am)=m(x-a{m}^2)$
$\Rightarrow y=mx-2am-a{m}^3$