#### Provide solution for rd sharma maths class 12 chapter 15 Tangents and Normals exercise 15.3 question 1 sub question 7

$\tan ^{-1} \frac{9}{13}$

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} &x^{2}=27 y \ldots \ldots(1) \\ &y^{2}=8 x \ldots \ldots(2) \end{aligned}

Considering second curve

\begin{aligned} &y^{2}=8 x \\ &x=\frac{y^{2}}{8} \end{aligned}

Substituting this in eq (1)

\begin{aligned} &\left(\frac{y^{2}}{8}\right)^{2}=27 y \\ &\frac{y^{4}}{64}=27 y \\ &y^{4}=y(27 \times 64) \end{aligned}

\begin{aligned} &y^{4}-y(27 \times 64)=0 \\ &y\left(y^{3}-(27 \times 64)\right)=0 \\ &y=0 \text { or } y^{3}=27 \times 64 \\ &y=0 \text { or } y=3 \times 4 \\ &y=0 \text { or } y=12 \end{aligned}

Now putting the value of y=0 & 12 in  $x=\frac{y^{2}}{8}$

When y = 0

$x^{2}=\frac{(0)^{2}}{8}=0$

When y = 12
\begin{aligned} &x^{2}=\frac{12^{2}}{8}=\frac{144}{8}=18 \\ &x=0 \text { or } x=18 \end{aligned}

Point of intersection are (0, 0) and (18, 12)

Since curves are $x^{2}=27 y \;\; \&\;\; y^{2}=8 x$

Differentiating above with respect to x
\begin{aligned} &\frac{d}{d x}\left(x^{n}\right)={ }^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0 \\ &=x^{2}=27 y \\ &=2 x=27 y \frac{d y}{d x} \\ &m_{1}=\frac{d y}{d x}=\frac{2 x}{27 y} \ldots \ldots \text { (3) } \end{aligned}

Now consider  $y^{2}=8 x$
\begin{aligned} &2 y \frac{d y}{d x}=\frac{8}{2 y} \\ &=m_{2}=\frac{d y}{d x}=\frac{4}{y} \end{aligned}

When (0,0)

\begin{aligned} &=m_{1}=\frac{d y}{d x}=\frac{2(0)}{27}=0 \\ &=m_{2}=\frac{d y}{d x}=\frac{4}{0}=\infty \end{aligned}

When (18, 12)

\begin{aligned} &=\frac{2 x}{27}=\frac{2 \times 18}{27}=\frac{4}{3} \\ &=m_{1}=\frac{4}{3} \\ &=m_{2}=\frac{d y}{d x}=\frac{8}{2 v}=\frac{8}{2 \times 12}=\frac{1}{3} \end{aligned}

Value of  m1=0 and $\frac{4}{3}$ , m2$\infty$ and $\frac{1}{3}$

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

When m1 is $\frac{4}{3}$ and m2 is $\frac{1}{3}$

\begin{aligned} &\operatorname{Tan} \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \\ &\operatorname{Tan} \theta=\left|\frac{\frac{4}{3}-\frac{1}{3}}{1+\left(\frac{4}{3}\right)\left(\frac{1}{3}\right)}\right| \\ &\operatorname{Tan} \theta=\left|\frac{\frac{3}{4}}{1+\frac{4}{9}}\right| \end{aligned}

\begin{aligned} &\operatorname{Tan} \theta=\left|\frac{3}{3} \times \frac{9}{13}\right| \\ &\operatorname{Tan} \theta=\left|\frac{9}{13}\right| \\ &\operatorname{Tan} \theta=\left(\frac{9}{13}\right) \\ &=\theta=\tan ^{-1} \frac{9}{13} \end{aligned}