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

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


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 )


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}

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