#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 12 Maths Textbook Solution.

ANSWER:$e^{2}\left ( x-y \right )=3$

HINTS:

First find the coordinates

GIVEN:

$y=xlog_{e}x$ Which is parallel  to $2x-2y+3=0$

SOLUTION:

$2x-2y+3=0$                        $Y=xlog_{e}x$

$m=1$                        Upon differentiation

\begin{aligned} &\frac{d y}{d x}=x \cdot \frac{1}{x}+\log x \cdot 1 \\ &\Rightarrow 1+\log x=-1 \\ &\therefore \log _{e} x=-2 \\ &\; \; \; \; \; \; \; \; \quad x=\frac{1}{e^{x}} \\ &\; \; \; \; \; \; \; \; \; \; \; y=\frac{1}{e^{2}} \log \left(\frac{1}{e^{2}}\right)=-\frac{1}{e^{2}} \log e^{2}=-\frac{2}{e^{2}} \end{aligned}

\begin{aligned} &\text { Coordinates }=\left[\frac{1}{e^{2}},-\frac{2}{e^{2}}\right] \\ &\therefore \text { Equation= } y+\frac{2}{e^{2}}=1\left(x-\frac{1}{e^{2}}\right) \\ &\; \; \; \; \; \; \; \; \; \; \; \: \: \: \: \: \: \: \: \: \: y \cdot e^{2}+2=x e^{2}-1 \\ &\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; e^{2}(x-y)=3 \end{aligned}