Question

Three normals to the parabola $y^2=x$ are drawn a point $(c,0)$, then :

• A. $c=\frac{1}{4}$
• B. $c=\frac{1}{2}$
• C. $c>\frac{1}{2}$
• D. None of these

Answer 1

Answer: (C)

The slope form of the normal equation to the parabola
$y^2=4ax$ is
$y=mx-2am-a{m}^3$
The slope form of the normal to the parabola $y^2=4ax$ is
$y=mx-2am-a{m}^3$
Since, the given curve is $y^2=x$
Here, $a=\frac{1}{4}$
$\therefore y=mx-\frac{1}{2}m-\frac{1}{4}{m}^3$
If it passes through $(c,0)$, then
$0=mc-\frac{1}{2}m-\frac{1}{4}{m}^3$
$\Rightarrow m=0$ or $c-\frac{1}{2}-\frac{1}{4}{m}^2=0$
$\Rightarrow m=\pm 2\sqrt{c-\frac{1}{2}}$
For three normal values of $m$, it should be real.
$\therefore c>\frac{1}{2}$.