#### Need solution for rd sharma maths class 12 chapter 15 Tangents and Normals exercise 15.3 question 5

Hence, prove two curves $2x=y^{2}$ and $2xy=k$ cut at right angle ,if $k^{2}=8$.

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

Given –  $2x=y^{2}---(1)$

$2xy=k---(2)$

Prove -

Two curves  cut at right angle ,if $k^{2}=8$

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

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

Second curve is $2xy=k$

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

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

Since  m1& m2 cuts orthogonally
\begin{aligned} &=\frac{1}{y} \times \frac{-y}{x}=-1 \\ &=\frac{-1}{x}=-1 \\ &=x=1 \end{aligned}

Now solving (1) & (2), we get

$= 2x=y^{2}$$2xy = k$

Substituting  $2x=y^{2}$ in $2xy = k$ ,we get
\begin{aligned} &=\left(y^{2}\right) y=k \\ &=y^{3}=k \\ &=y=k^{\frac{1}{3}} \quad\left\{:: a^{m}=n, a=n^{\frac{1}{m}}\right\} \end{aligned}

Substituting  $y=k^{\frac{1}{3}}$      in $2x=y^{2}$ and x=1  we get

$=2x=\left (k^{\frac{1}{3}} \right )^{2}$
\begin{aligned} &=2 \times 1=k^{\frac{2}{3}} \\ &=2=k^{\frac{2}{3}} \end{aligned}

\begin{aligned} &=k^{\frac{2}{3}}=2 \end{aligned}   ,  cube both sides

\begin{aligned} &=\left (k^{\frac{2}{3}} \right )^{3}=2^{3}\\ &=k^{2}=8 \end{aligned}

Hence, prove two curves cut at right angles if \begin{aligned} &k^{2}=8 \end{aligned}