If the tangent of the curve, $\mathrm{y}=\mathrm{e}^{\mathrm{x}}$ at a point $\left(c, e^{c}\right)$ and the normal to the parabola, $y^{2}=4 x$ at the point $(1,2)$ intersect at the same point on the $x$-axis, then the value of $c$ is_______________
$y=e^{x} \Rightarrow \frac{d y}{d x}=e^{x}$
$\mathrm{m}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\left(\mathrm{c}, \mathrm{e}^{\mathrm{c}}\right)}=\mathrm{e}^{\mathrm{c}}$
$\Rightarrow \quad$ Tangent at $\left(\mathrm{c}, \mathrm{e}^{\mathrm{c}}\right)$
$y-e^{c}=e^{c}(x-c)$
it intersect $x$-axis
Put $\quad y=0 \Rightarrow x=c-1$....(1)
Now $\mathrm{y}^{2}=4 \mathrm{x} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2}{\mathrm{y}} \Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(1,2)}=1$
$\Rightarrow$ Slope of normal $=-1$
Equation of normal $y-2=-1(x-1)$
$x+y=3$ it intersect $x$-axis
Put $\quad y=0 \Rightarrow x=3$ ...(2)
Points are same
$\Rightarrow \quad \mathrm{x}=\mathrm{c}-1=3$
$\Rightarrow \quad \mathrm{c}=4$