Question:
If the tangent to the curve, $y=e^{x}$ 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
Solution:
For $(1,2)$ of $y^{2}=4 x \Rightarrow t=1, a=1$
Equation of normal to the parabola
$\Rightarrow t x+y=2 a t+a t^{3}$
$\Rightarrow x+y=3$ intersect $x$-axis at $(3,0)$
$y=e^{x} \Rightarrow \frac{d y}{d x}=e^{x}$
Equation of tangent to the curve
$\Rightarrow y-e^{c}=e^{c}(x-c)$
$\because$ Tangent to the curve and normal to the parabola
intersect at same point.
$\therefore 0-e^{c}=e^{c}(3-c) \Rightarrow c=4$