For the parabola $y^2=4x$, $a=1$

Let $\left(a{t}^2, 2at\right)=\left(1, 2\right)$

$\Rightarrow t=1$

So, equation of normal is $tx+y=2at+a{t}^3$.

$\Rightarrow x+y=3$ intersects $x-$axis at $\left(3, 0\right)$

Given curve $y=e^x\Rightarrow \frac{dy}{dx}=e^x$

So, equation of tangent to the curve at $\left(c, e^c\right)$ is $y-e^c=e^c\left(x-c\right)$.

At $\left(3, 0\right)\Rightarrow 0-e^c=e^c\left(3-c\right)\Rightarrow c=4$.