Question

If the tangent to the curve, $\mathrm{y}={\mathrm{e}}^{\mathrm{x}}$ at a point $\left(\mathrm{c},{\mathrm{e}}^{\mathrm{c}}\right)$ and the normal to the parabola,${\mathrm{y}}^2=4\mathrm{x}$ at the point $\left(1,2\right)$ intersect at the same point on the $x$-axis, then the value of $c$ is

Answer 1

Step 1: Finding the relation between $x$ and $c$:

$\begin{array}{c}\mathrm{y}={\mathrm{e}}^{\mathrm{x}}\end{array}$

Differentiate this with respect to $x$ to get the slope,

$\begin{array}{rcl}\frac{dy}{dx}& =& e^x\\ & \Rightarrow & \mathrm{m}={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left(\mathrm{c},{\mathrm{e}}^{\mathrm{c}}\right)}\\ & =& {\mathrm{e}}^{\mathrm{c}}\end{array}$

Equation of tangent at point $\left(\mathrm{c},{\mathrm{e}}^{\mathrm{c}}\right)$,

$\mathrm{y}-{\mathrm{e}}^{\mathrm{c}}={\mathrm{e}}^{\mathrm{c}}\left(\mathrm{x}-\mathrm{c}\right)\begin{array}{rcl}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{y}\mathbf{-}{\mathbf{y}}_{\mathbf{1}}& \mathbf{=}& \mathbf{m}\left(x-x_1\right)\mathbf{]}\end{array}$

It crosses the $x$-axis

So, substitute $\mathrm{y}=0$ in the above equation:

$\mathrm{x}=\mathrm{c}-1...\left(\mathrm{i}\right)$

Step 2: Finding the value of $x$:

Now,

$y^2=4x$

Differentiate the above equation with respect to $x$ to get the slope,

$\begin{array}{rcl}2y\frac{dy}{dx}& =& 4\\ & \Rightarrow & m={\frac{dy}{dx}}_{\left(1.2\right)}\\ & =& \frac{2}{2}\\ & =& 1\end{array}$

The slope of normal$=-\frac{1}{m}$.

$=-1$

The equation of normal is

$\begin{array}{rcl}\mathrm{y}-2& =& -1\left(\mathrm{x}-1\right)\left[\because y-y_1=m\left(x-x_1\right)\right]\\ & \Rightarrow & x+y=3\end{array}$

It crosses the $x$ axis .So, substitute $\mathrm{y}=0$ in the above equation,

$\begin{array}{rcl}x+0& =& 3\\ & \Rightarrow & x=3\end{array}$

Step 3: Find the value of $c$:

By substituting the value of $x$ in the equation $\left(\mathrm{i}\right)$ as tangent and normal meet at same point, we get

$\begin{array}{rcl}3& =& c-1\\ & \Rightarrow & c=4\end{array}$

Hence, the value of $\mathrm{c}$ is $4$.