#### Need solution for rd sharma maths class 12 chapter 15 Tangents and Normals exercise 15.3 question 4

Hence, prove two curves $4x=y^{2}$ and $4xy=k$ cut at right angle ,if $k^{2}=512$.

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

Given –  $4x=y^{2}---(1)$

$4xy=k---(2)$

Prove -

Two curves  cut at right angle ,if $k^{2}=512$

Consider First curve is $4x=y^{2}$

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

Second curve is $4xy=k$

Differentiating above with respect to x,
\begin{aligned} &=4\left(y+x \frac{d y}{d x}\right)=0 \\ &=\left(y+x \frac{d y}{d x}\right)=0 \\ &=\frac{x d y}{d x}=-y=\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{2}{y} \times \frac{-y}{x}=-1 \\ &=\frac{-2}{x}=-1 \\ &=x=2 \end{aligned}

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

$= 4xy = k$  &  $4x=y^{2}$

Substituting  $4x=y^{2}$ in $4xy=k$ ,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 $4x=y^{2}$ we get

$=4x=\left (k^{\frac{1}{3}} \right )$    and put x=2
\begin{aligned} &=4 \times 2=k^{\frac{2}{3}} \\ &=8=k^{\frac{2}{3}} \end{aligned}

\begin{aligned} &=k^{\frac{2}{3}}=8 \end{aligned}   ,  cube both sides

\begin{aligned} &=\left (k^{\frac{2}{3}} \right )^{3}=8^{3}\\ &=k^{2}=512 \end{aligned}