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

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

Answers (1)

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}

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