#### Please Solve RD Sharma Class 12 Chapter15 Tangents and Normals Exercise Fill in the blanks Question 7 Maths Textbook Solution.

$\frac{\pi}{2}$

Hint:

First we need to find their point of intersection. So solve these given equations.

Given:

Given curves,

$y= x^{2}$    and  $x= y^{2}$

To find:

We have to find the angle of intersection of the given curves at  $\left ( 0,0 \right )$

Solution:

Here      $y= x^{2}$                                                                                                                                                                 … (i)

$x= y^{2}$                                                                                                                                                              … (ii)

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

$\begin{array}{ll} \Rightarrow & x^{4}=x \\\\ \Rightarrow & x^3(x-1)=0 \end{array}$

Which gives   $x= 0,x= 1$

Therefore points of intersection of the curves are  $\left ( 0,0 \right )$ and $\left ( 1,1 \right )$

On differentiating equation (i) and (ii), we get

$\Rightarrow \quad \frac{d y}{d x}=2 x$

Similarly, $2 y \frac{d y}{d x}=1$

$\Rightarrow \quad \frac{d y}{d x}=\frac{1}{2 y}$

To find the angle of intersection at $\left ( 0,0 \right )$

So, $m_{1}$ slope of tangent to curve $y= x^{2}$ at  $\left ( 0,0 \right )$

$= 2x= 0$

So tangent is parallel to $x-$ axis.

$m_{2}$ = slope of tangent to curve  $x= y^{2}$  at $\left ( 0,0 \right )$

$= \frac{1}{2y}$  which is not defined.

So tangent is parallel to $y-$ axis.

Now, one tangent is parallel to  $x-$axis and other is parallel to  $y-$axis.

Hence angle between tangents is right angle. i.e. $\frac{\pi}{2}$