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#### Provide solution for rd sharma maths class 12 chapter 15 Tangents and Normals exercise 15.3 question 1 sub question 9

$\tan^{-1}\frac{4\sqrt{2}}{7}$

Hint - The angle of intersection of curves is$\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$

m1=slope of first curve.   m2=slope of second curve.
Given –
\begin{aligned} &y=4-x^{2} \ldots \ldots(1) \\ &y=x^{2} \ldots \ldots(2) \end{aligned}

Substituting eq (2) in (1) we get

,\begin{aligned} &x^{2}=4-x^{2} \\ &2 x^{2}=4 \\ &x^{2}=2 \\ &x=\pm \sqrt{2} \end{aligned}

Put \begin{aligned} &x=\pm \sqrt{2} \end{aligned} in eq (2), we get
When \begin{aligned} &x=\sqrt{2} \end{aligned}

\begin{aligned} &y=(\sqrt{2})^{2}=2 \\ &y=2 \end{aligned}

When \begin{aligned} &x=-\sqrt{2} \end{aligned}

\begin{aligned} &y=(-\sqrt{2})^{2}=2 \\ &y=2 \end{aligned}

Thus two curves intersect at \begin{aligned} &\left (\sqrt{2},2 \right ) \end{aligned}and \begin{aligned} &\left (-\sqrt{2},2 \right ) \end{aligned}

Since curves are  $y=4-x^{2}$ and $y=x^{2}$

Differentiating above with respect to x

As we know, $\frac{d}{d x}\left(x^{n}\right)={ }^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0$

Consider first curve

\begin{aligned} &=y=4-x^{2} \\ &=\frac{d y}{d x}=0-2 x \\ &m_{1}=\frac{d y}{d x}=-2 x \end{aligned}

Now consider second curve  $y=x^{2}$

$=m_{2}=\frac{d y}{d x}=2 x$

When \begin{aligned} &\left (\sqrt{2},2 \right ) \end{aligned}
\begin{aligned} &=m_{1}=\frac{d y}{d x}=-2 x=-2(\sqrt{2})=-2 \sqrt{2} \\ &=m_{1}=-2 \sqrt{2} \end{aligned}

When \begin{aligned} &\left (-\sqrt{2},2 \right ) \end{aligned}

\begin{aligned} &=m_{1}=\frac{d y}{d x}=2 x=2(-\sqrt{2})=-2 \sqrt{2} \\ \end{aligned}

Angle of intersection of two curves is given by $\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
\begin{aligned} &\operatorname{Tan} \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \\ &\operatorname{Tan} \theta=\left|\frac{2 \sqrt{2}-(-2 \sqrt{2})}{1+(2 \sqrt{2})(-2 \sqrt{2})}\right| \\ &\operatorname{Tan} \theta=\left|\frac{2 \sqrt{2}+2 \sqrt{2}}{1-(4)(2)}\right|=\left|\frac{4 \sqrt{2}}{1-8}\right|=\left|\frac{4 \sqrt{2}}{7}\right| \\ &=\theta=\tan ^{-1} \frac{4 \sqrt{2}}{7} \end{aligned}