#### please solve rd sharma class 12 chapter 15 Tangents and Normals exercise 15.3 , question 1 sub question 2 maths textbook solution

$\theta=\tan ^{-1}\left(\frac{9}{2}\right)$

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- $y=x^{2} \ldots \ldots(1) \;\; \& \;\; x^{2}+y^{2}=20 \ldots \ldots(2)$

First curve is $y=x^{2}$

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

\begin{aligned} &=\frac{d y}{d x}=2 x \\ &=m_{1}=2 x \ldots \ldots(3) \end{aligned}

Second curve is $x^{2}+y^{2}=20$

Differentiating above with respect to x,

\begin{aligned} &=2 x+2 y \frac{d y}{d x}=0 \\ &=2 y \frac{d y}{d x}=-2 x \end{aligned}

\begin{aligned} &=m_{2}=\frac{d y}{d x}=\frac{-2 x}{2 y}=\frac{-x}{y} \\ &=m_{2}=\frac{-x}{y} \end{aligned}

Substituting (1) in (2),we get

\begin{aligned} &=y=x^{2} \\ &=y+y^{2}=20 \\ &=y^{2}+y-20=0 \end{aligned}

By factorization method

\begin{aligned} &=y^{2}+y-20=0 \\ &=y^{2}+5 y-4 y-20=0 \\ &=y(y+5)-4(y+5)=0 \\ &=(y-4)(y+5)=0 \end{aligned}

=y=-5 and y=4

Substituting  y=-5 and y=4 in (1)

When, y = -5

$x^{2}=-5$  This is not possible

When y=4

$x^{2}=4\\ x=\pm 2$

Substituting the values  for m1& m2 , we get,
\begin{aligned} &=m_{1}=\frac{d y}{d x}=2x \end{aligned}

When, x=2

$=m_{1}=2(2)=4$

When x=-2

$=m_{1}=2(-2)=-4$

Value of m1 is 4, -4

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

When y=4 & x=2

$=m_{2}=\frac{d y}{d x}=\frac{-x}{y}=\frac{-2}{4}=\frac{-1}{2}$

When y=4 & x=-2

$=m_{2}=\frac{d y}{d x}=\frac{-x}{y}=\frac{-(-2)}{4}=\frac{1}{2}$

Values of m2 is $\frac{1}{2}$ and $\frac{-1}{2}$

As we know, Angle of intersection of two curves is given by $\operatorname{tan} \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$

When m1 is 4 and m2 is $\frac{-1}{2}$

Then,

\begin{aligned} &\operatorname{Tan} \theta=\left(\frac{4-\left(\frac{-1}{2}\right)}{1+4 \times\left(\frac{-1}{2}\right)}\right) \\ &\operatorname{Tan} \theta=\left(\frac{\left(\frac{9}{2}\right)}{-1}\right) \\ &\tan \theta=\left|\left(\frac{-9}{2}\right)\right| \\ &\tan \theta=\frac{9}{2} \\ &\theta=\tan ^{-1}\left(\frac{9}{2}\right) \end{aligned}

When $m_{1}=-4$   and  $m_{2}=\frac{1}{2}$

\begin{aligned} &\operatorname{Tan} \theta=\left|\frac{\left(-4-\frac{1}{2}\right)}{1+(-4)\left(\frac{1}{2}\right)}\right|=\left|\frac{\left(\frac{-8-1}{2}\right)}{1-2}\right|=\left|\frac{\frac{-9}{2}}{-1}\right| \\ &\operatorname{Tan} \theta=\left|\frac{9}{2}\right| \\ &\operatorname{Tan} \theta=\frac{9}{2} \\ &=\theta=\tan ^{-1}\left(\frac{9}{2}\right) \end{aligned}

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