#### Provide solution for RD Sharma maths class 12 Chapter 15 Tangents and Normals Exercise Very Short Answer Question 8 maths text book solution.

Answer : $(x, y)=\left(\frac{1}{4}, \frac{1}{2}\right)$

Hint : $\text { slope of tangent }=\tan \frac{\pi}{4}=1$

Given :

Here the given curve, $y^{2}=x$ where the tangent line makes an angle $\frac{\pi }{4}$ with x- axis.

We have to find the co-ordinate of the point on the given curve.

Solution:

Let the required point be $(x,y)$.

We know,

The tangent makes an angle $45^{o}$with the x-axis.

$\therefore$              $\text { slope of tangent }=\tan 45^{\circ}=1$

Since the point lies on the curve

Hence,  $y_{1}^{2}=x_{1}$

Now,

$y^{2}=x$

Differentiating with respect to x,

$\begin{array}{ll} \Rightarrow \quad & 2 y \frac{d y}{d x}=1 \\ \Rightarrow & \frac{d y}{d x}=\frac{1}{2 y} \end{array}$

$\therefore \quad \text { slope of tangent }=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{1}{2 y_{1}}$

$\Rightarrow \quad \frac{1}{2 y_{1}}=1 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \text { slope of tangent at angle } \frac{\pi}{4}=\tan \frac{\pi}{4}=1\right]$

$\Rightarrow \quad 2 y_{1}=1$

$\Rightarrow \quad y_{1}=\frac{1}{2}$

Now, we have,

\begin{aligned} &x_{1}=y_{1}^{2} \\ &\Rightarrow x_{1}=\left(\frac{1}{2}\right)^{2} \\ &\Rightarrow \quad x_{1}=\frac{1}{4} \end{aligned}

Hence, $(x, y)=\left(\frac{1}{4}, \frac{1}{2}\right)$