Given curve is $y^2=x ...\left(1\right)$

$y^2=4·\frac{1}{4}x$

Equation $\left(1\right)$ represents a parabola, where $a=\frac{1}{4}$. The equation of normal to a parabola is

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

$\Rightarrow y=mx-2\left(\frac{1}{4}\right)m-\left(\frac{1}{4}\right)m^3$

$\Rightarrow y=mx-\frac{m}{2}-\frac{m^3}{4}$

The normal passes through $(c, 0)$

$\Rightarrow 0=m\left(c\right)-\frac{m}{2}-\frac{m^3}{4}$

$\Rightarrow 0=m\left(c-\frac{1}{2}-\frac{m^2}{4}\right)$

$\Rightarrow m=0$ or $c-\frac{1}{2}-\frac{m^2}{4}=0$

$\Rightarrow \frac{m^2}{4}=c-\frac{1}{2}$

$\Rightarrow m^2=4c-2$

$\Rightarrow m=\pm \sqrt{4c-2}$

From data, the normal are perpendicular to each other. So, 

$\left(\sqrt{4c-2}\right)\left(-\sqrt{4c-2}\right)=-1$

$\Rightarrow 4c-2=1$

$\Rightarrow c=\frac{3}{4}$