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

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

Answers (1)

Answer:

                \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)

Posted by

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