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

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

Answers (1)

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}

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