#### please solve rd sharma class 12 chapter 15 Tangents and Normals exercise 15.3 , question 1 sub question 3 maths textbook solution

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

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- $2 y^{2}=x^{3} \ldots \ldots(1)\;\; \& \;\; y^{2}=32 x \ldots \ldots(2)$

First curve is $2y=x^{3}$

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} &=4 y \frac{d y}{d x}=3 x^{2} \\ &=m_{1}=\frac{d y}{d x}=\frac{3 x^{2}}{4 y} \ldots \ldots \text { (3) } \end{aligned}

Second curve is $y^{2}=32x$

Differentiating above with respect to x,

\begin{aligned} &=2 y \frac{d y}{d x}=32 \\ &=\frac{d y}{d x}=\frac{32}{2 y}=\frac{16}{y} \\ &=m_{2}=\frac{d y}{d x}=\frac{16}{y} \ldots \ldots \text { (4) } \end{aligned}

Substituting (1) in (2),we get

\begin{aligned} &=2 y^{2}=x^{3} \\ &=2(32 x)=x^{3} \\ &=64 x=x^{3} \\ &=x^{3}-64 x=0 \end{aligned}

\begin{aligned} &=x\left(x^{2}-64\right)=0 \\ &=x=0 \text { or } x^{2}-64=0 \\ &=x=0 \text { or } x^{2}=64 \\ &=x=0 \& x=\pm 8 \end{aligned}

Substituting  x=0 or $\pm 8$ in (2)

$y^{2}=32x$  This is not possible

When x=0

\begin{aligned} &y^{2}=32(0) \\ &y=0 \end{aligned}

When x=8

\begin{aligned} &y^{2}=32(8) \\ &y^{2}=256 \\&y=\pm 16 \end{aligned}

Substituting the values  for m1& m2 , we get,
When, x=0

\begin{aligned} &=m_{1}=\frac{d y}{d x}=\frac{3 x^{2}}{4 y} \\ &=m_{1}=\frac{3(0)}{4(16)}=0 \end{aligned}

When x=8,y=16

$=m_{1}=\frac{3(8)^{2}}{4(16)}=3$

Value of m1 is 0 and 3

When x=0 & y=0

\begin{aligned} &=m_{2}=\frac{d y}{d x}=\frac{16}{y} \\ &=m_{2}=\frac{16}{0}=\infty \quad\left\{:: \frac{1}{0}=\infty\right\} \end{aligned}

When y=16

Values of m2 is $\infty$ and 1

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 0 and m2 is $\infty$

$\theta=\tan ^{-1}(\infty)$ As we know $\tan \left(\frac{\pi}{2}\right)=\infty$
\begin{aligned} &\operatorname{Tan} \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \\ &\operatorname{Tan} \theta=\left(\frac{\infty-0}{1+\infty \times 0}\right) \\ &\operatorname{Tan} \theta=\infty \\ &\theta=\tan ^{-1}(\infty) \\ &=\tan ^{-1}(\infty)=\frac{\pi}{2} \\ &\theta=\frac{\pi}{2} \end{aligned}

$\infty=\frac{\pi}{2}$

When $m_{1}=3$   and  $m_{2}=1$

\begin{aligned} &\operatorname{Tan} \theta=\left|\frac{(3-1)}{1+(1)(3)}\right|=\left|\frac{2}{1+3}\right|=\left|\frac{2}{4}\right| \\ &\operatorname{Tan} \theta=\left|\frac{1}{2}\right| \\ &\operatorname{Tan} \theta=\frac{1}{2} \\ &=\theta=\tan ^{-1}\left(\frac{1}{2}\right) \end{aligned}

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

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- $2 y^{2}=x^{3} \ldots \ldots(1)\;\; \&\;\; y^{2}=32 x \ldots \ldots(2)$

First curve is $2y^{2}=x^{3}$

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} &=4 y \frac{d y}{d x}=3 x^{2} \\ &=m_{1}=\frac{d y}{d x}=\frac{3 x^{2}}{4 y} \ldots \ldots \text { (3) } \end{aligned}

Second curve is $y^{2}=32x$

Differentiating above with respect to x,

\begin{aligned} &=2 y \frac{d y}{d x}=32 \\ &=\frac{d y}{d x}=\frac{32}{2 y}=\frac{16}{y} \\ &=m_{2}=\frac{d y}{d x}=\frac{16}{y} \ldots \ldots \text { (4) } \end{aligned}

Substituting (1) in (2),we get

\begin{aligned} &=2 y^{2}=x^{3} \\ &=2(32 x)=x^{3} \\ &=64 x=x^{3} \\ &=x^{3}-64 x=0 \end{aligned}

\begin{aligned} &=x\left(x^{2}-64\right)=0 \\ &=x=0 \text { or } x^{2}-64=0 \\ &=x=0 \text { or } x^{2}=64 \\ &=x=0 \& x=\pm 8 \end{aligned}

Substituting  \begin{aligned} x=0 \;\; \& \;\; x=\pm 8 \end{aligned}   in (1)

\begin{aligned} y^{2}=32x \end{aligned}

When, x=0

\begin{aligned} &y^{2}=32(0) \\ &y=0 \end{aligned}

When x=8

\begin{aligned} &y^{2}=32(8) \\ &y^{2}=256 \\ &y=\pm 16 \end{aligned}

Substituting the values  for m1& m2 , we get,

when,x=0
\begin{aligned} &=m_{1}=\frac{d y}{d x}=\frac{3 x^{2}}{4 y} \\ &=m_{1}=\frac{3(0)}{4(16)}=0 \end{aligned}

When, x=8, y=16

$=m_{1}=\frac{3(8)^{2}}{4(16)}=3$

Value of m1 is 0 and 3.

When x=-0,y=0

\begin{aligned} &=m_{2}=\frac{d y}{d x}=\frac{16}{y} \\ &=m_{2}=\frac{16}{0}=\infty \quad\left\{:: \frac{1}{0}=\infty\right\} \end{aligned}

when y=16

$=m_{2}=\frac{16}{y}=\frac{16}{16}=1$

Value of m2 is  $\infty$  and 1

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 0 and m2 is $\infty$

\begin{aligned} &\operatorname{Tan} \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \\ &\operatorname{Tan} \theta=\left(\frac{\infty-0}{1+\infty \times 0}\right) \\ &\operatorname{Tan} \theta=\infty \\ &\theta=\tan ^{-1}(\infty) \\ &=\tan ^{-1}(\infty)=\frac{\pi}{2} \\ &\theta=\frac{\pi}{2} \end{aligned}

When $m_{1}=3$   and  $m_{2}=1$

\begin{aligned} &\operatorname{Tan} \theta=\left|\frac{(3-1)}{1+(1)(3)}\right|=\left|\frac{2}{1+3}\right|=\left|\frac{2}{4}\right| \\ &\operatorname{Tan} \theta=\left|\frac{1}{2}\right| \\ &\operatorname{Tan} \theta=\frac{1}{2} \\ &=\theta=\tan ^{-1}\left(\frac{1}{2}\right) \end{aligned}