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

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

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