Question

Three normals to the parabola $y^2=x$ are drawn through 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

The correct option is C $C>\frac{1}{2}$

The slope form of the normal to the parabola $y^2=4ax$ is $y=mx-2am-a{m}^3$.
For the given curve $y^2=x$, we will have $4a=1\Rightarrow a=\frac{1}{4}$
Hence the equation of the normal is $y^2=mx-\frac{1}{2}m-\frac{1}{4}{m}^3$
If it passes through , 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 normals, value of m should be real, $\therefore C>\frac{1}{2}$.