A curve C passes through origin and has the property that at each point  $\left(\mathrm{x}, \mathrm{y}\right)$ on it the normal line at that point passes through $\left(1, 0\right)$. The equation of a common tangent to the curve C and the parabola  ${\mathrm{y}}^2=4\mathrm{x}$ is

Slope of the normal$=\frac{\mathrm{y}}{\mathrm{x}-1}$

$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1-\mathrm{x}}{\mathrm{y}}\\ \frac{{\mathrm{y}}^2}{2}=\mathrm{x}-\frac{{\mathrm{x}}^2}{2}+\mathrm{C}\end{array}$

II) Passes through $\left(1, 0\right)$ hence $\mathrm{C}=0{\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}=0$

Now, tangent to ${\mathrm{y}}^2=4\mathrm{x}$

$\mathrm{y}=\mathrm{mx}+\frac{1}{\mathrm{m}}$

If it touches the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}=0$

$\left|\frac{\mathrm{m}+(1/\mathrm{m})}{\sqrt{1+{\mathrm{m}}^2}}\right|=1\Rightarrow 1+{\mathrm{m}}^2={\mathrm{m}}^2\Rightarrow \mathrm{m}\to \mathrm{\infty }$

Then, tangent is y-axis i.e., x = 0