# If the tangent of the curve,

Question:

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_______________

Solution:

$y=e^{x} \Rightarrow \frac{d y}{d x}=e^{x}$

$\mathrm{m}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\left(\mathrm{c}, \mathrm{e}^{\mathrm{c}}\right)}=\mathrm{e}^{\mathrm{c}}$

$\Rightarrow \quad$ 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

$\Rightarrow \quad \mathrm{x}=\mathrm{c}-1=3$

$\Rightarrow \quad \mathrm{c}=4$