#### please solve rd sharma class 12 chapter 15 Tangents and Normals exercise 15.3 , question 1 sub question 6 maths textbook solution

$\theta=\tan^{-1}3$

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}+4 y^{2}=8 \ldots \ldots(1) \\ &x^{2}-2 y^{2}=2 \ldots \ldots \text { (2) } \end{aligned}

Considering second curve

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

Substituting this in eq (1)

\begin{aligned} &2+2 y^{2}+4 y^{2}=8 \\ &6 y^{2}=6 \\ &y^{2}=1 \\ &y=\pm 1 \end{aligned}

Now putting the value of y in  $x^{2}=2+2 y^{2}$

When y = 1

\begin{aligned} &x^{2}=2+2(1)^{2} \\ &x^{2}=2+2 \\ &x^{2}=4 \\ &x=\pm 2 \end{aligned}

Point of intersection are (2,1) and (-2,-1)

Since curves are $x^{2}+4 y^{2}=8\;\; \&\;\; x^{2}-2 y^{2}=2$

Differentiating above with respect to x

\begin{aligned} &x^{2}+4 y^{2}=8 \\ &2 x+8 y \frac{d y}{d x}=0 \\ &8 y \frac{d y}{d x}=-2 x \\ &=\frac{d y}{d x}=\frac{-2 x}{8 y} \end{aligned}

\begin{aligned} &m_{1}=\frac{d y}{d x}=\frac{-x}{4 y} \ldots \ldots(3) \\ &\frac{d}{d x}\left(x^{n}\right)={ }^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0 \\ &=x^{2}-2 y^{2}=2 \end{aligned}

\begin{aligned} &=2 x-\frac{4 y(d y)}{d x}=0 \\ &=2 x=4 y \frac{d y}{d x} \\ &=\frac{d y}{d x}=\frac{2 x}{4 y} \\ &m_{2}=\frac{d y}{d x}=\frac{x}{2 y} \ldots \ldots \text { (4) } \end{aligned}

When (2,1)

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

When (-2,-1)

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

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

When m1is $\frac{-1}{2}$ and m2 is 1
\begin{aligned} &\operatorname{tan} \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \\ &\operatorname{tan} \theta=\left|\frac{-\frac{1}{2}-1}{1+\left(-\frac{1}{2}\right)(1)}\right| \end{aligned}

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