#### Provide solution for rd sharma maths class 12 chapter 15 Tangents and Normals exercise 15.3 question 3 sub question 3

$m_{1} \times m_{2}=-1$

Hence, two
curves intersect orthogonally.

Hint - Two curves intersects orthogonally if $m_{1} \times m_{2}=-1$, where m1 and m2 are the slopes of two curves.

Given –  $y^{2}=8x-------(1)$

$2x^{2}+y^{2}=10-------(2)$

The point of intersection of two curve $(1,2\sqrt{2})$.

First curve is $y^{2}=8x$

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} &=2 y \frac{d y}{d x}=8 \\ &=\frac{d y}{d x}=\frac{8}{2 y} \\ &=m_{1}=\frac{4}{y}---(3) \end{aligned}
Second curve is $2x^{2}+y^{2}=10$

Differentiating above with respect to x,
\begin{aligned} &=4 x+\frac{2 y d y}{d x}=0 \\ &=2 y \frac{d y}{d x}=-4 x \\ &=\frac{d y}{d x}=\frac{-4 x}{2 y}=\frac{-2 x}{y} \\ &=m_{2}=\frac{-2 x}{y}---(4) \end{aligned}

Substituting $(1,2\sqrt{2})$ for m1& m2 , we get,
\begin{aligned} &=m_{1}=\frac{4}{y}=\frac{4}{2 \sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2} \\ &=m_{1}=\sqrt{2} \\ &=m_{2}=\frac{-2 x}{y}=\frac{-2 \times 1}{2 \sqrt{2}}=\frac{-1}{\sqrt{2}} \\ &=m_{2}=\frac{-1}{\sqrt{2}} \end{aligned}

When $m_{1}=\sqrt{2}$ and  m_{2}=\begin{aligned}\frac{-1}{\sqrt{2}} \end{aligned}

Two curves intersects orthogonally if $m_{1} \times m_{2}=-1$

\begin{aligned}=\sqrt{2} \times \frac{-1}{\sqrt{2}}=-1 \end{aligned}

Hence, two curves $y^{2}=8x \;\;\;\& \;\;\; 2x^{2}+y^{2}=10$ intersect orthogonally.