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

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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)


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}

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