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Need solution for RD Sharma Maths Class 12 Chapter 15 Tangents and Normals Excercise Fill in the blanks Question 18

Answers (1)

Answer:

$\left ( \frac{1}{4} ,\frac{1}{2}\right )$

Hint:

First find the slope of curve $y^{2}=x$ , then compare with slope $45^{\circ}$ with x-axis.

i.e. slope $\tan\frac{\pi}{4}=1$

Given:

Given curve, $y^{2}=x$ , the tangent at which makes an angle of $45^{\circ}$ with x-axis.

To find:

We have to find the point on the given curve.

Solution:

We have

$y^{2}=x$ … (i)

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

$\Rightarrow$ $2y\frac{dy}{dx}=1$ $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$

$\Rightarrow$ $\frac{dy}{dx}=\frac{1}{2y}$ … (ii)

Also tangent makes an angle of $45^{\circ}\left ( \frac{\pi}{4} \right )$ with x-axis

$\Rightarrow$ $\frac{dy}{dx}=\tan\left ( \frac{\pi}{4} \right )$ $\left[\because \tan \left(\frac{\pi}{4}\right)=1\right]$

$\Rightarrow$ $\frac{dy}{dx}=1$ … (iii)

From equation (ii) and (iii), we get

$\begin{aligned} &\Rightarrow \quad \frac{1}{2 y}=1 \\ &\Rightarrow \quad y=\frac{1}{2} \end{aligned}$

Put value of $y$ in equation $y^{2}=x$

$\Rightarrow$ $x=\frac{1}{4}$

Hence required point is $\left ( \frac{1}{4} ,\frac{1}{2}\right )$

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