#### Provide Solution for RD Sharma Class 12 Chapter 15 Tangents and Normals  Exercise Fill in the blanks  Question 4

$\left(e^{-2},-2 e^{-2}\right)$

Hint:

1. Slope of normal  $\frac{-1}{\frac{dy}{dx}}$
2. Slope of line is  $m$ . i.e.  $y=mx+c$

Given:

Given curve,  $y=x \log _{e} x$  and the line  $2x-2y=3$

To find:

We have to find the co-ordinates of a point on the given curve at which the normal is parallel to the line  $2x-2y=3$

Solution:

We have,

$y=x \log _{e} x$                                                                                                                                      …(i)

Differentiating both side with respect to $x$ , we get

$\Rightarrow \quad \frac{d y}{d x}=x \frac{d(\log x)}{d x}+\log x \frac{d(x)}{d x}$                                                                $\left[\because \frac{d(u v)}{d x}=u \frac{d v}{d x}+\frac{d u}{d x} v\right]$

$\Rightarrow \quad \frac{d y}{d x}=x \cdot \frac{1}{x}+\log x(1)$                                                                                    $\left[\begin{array}{l} \because \frac{d(\log x)}{d x}=\frac{1}{x} \\ \because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1} \end{array}\right]$

$\Rightarrow \quad \frac{d y}{d x}=1+\log x$

The Slope of normal   $= \frac{-1}{\frac{dy}{dx}}$

$= \frac{-1}{1+\log x}$

The line, $2x-2y=3$

\begin{aligned} &\Rightarrow \quad 2 y=2 x-3 \\ &\Rightarrow \quad y=x-\frac{3}{2} \end{aligned}

Comparing this equation with the formula  $y=mx+c$ , we can get the slope $m$ .

Here the slope of line   $2x-2y=3$  is  $1$

As the slope of parallel lines are equal,

Therefore,  $\frac{-1}{1+\log x}=1$

$\begin{array}{ll} \Rightarrow & \log x=-2 \\ \Rightarrow & x=e^{-2} \end{array}$

Putting the value of $x$ in equation (i)

$\Rightarrow \quad y=-2 e^{-2}$

Therefore the coordinate of the point is  $\left(e^{-2},-2 e^{-2}\right)$