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  • Need solution for rd sharma maths class 12 chapter 15 Tangents and Normals exercise 15.3 question 6

Need solution for rd sharma maths class 12 chapter 15 Tangents and Normals exercise 15.3 question 6

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Hence, prove two curves xy=4 and x^{2}+y^{2}=8 touch each other

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

Given –  xy=4---(1)

              x^{2}+y^{2}=8---(2)

Prove - 

Consider First curve is xy=4
x=\frac{4}{y}

Substituting  x=\frac{4}{y}  in eq (2), we get
\begin{aligned} &=\left(\frac{4}{y}\right)^{2}+y^{2}=8 \\ &=\frac{16}{y^{2}}+y^{2}=8 \\ &=\frac{16+y^{2}\left(y^{2}\right)}{y^{2}}=8 \end{aligned}

\begin{aligned} &=16+y^{4}=8 y^{2} \\ &=y^{4}-8 y^{2}+16=0 \\ &=\left(y^{2}-4\right)^{2}=0 \quad\left\{::(a-b)^{2}=a^{2}+b^{2}+2 a b\right\} \\ &=y^{2}=4 \\ &=y=\pm 2 \end{aligned}
Now substituting value of \begin{aligned}y=\pm 2 \end{aligned}  in eq \begin{aligned} x=\frac{y}{4} \end{aligned}

When y = 2

\begin{aligned} x=\frac{4}{2}=2 \end{aligned}

When y = -2

\begin{aligned} x=\frac{4}{-2}=-2 \end{aligned}

Thus, two curves intersect at (2, 2) and (-2,-2)

Consider first curve xy = 4

Differentiating above with respect to y,

As we know, \frac{d}{d x}\left(x^{n}\right)=^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0
\begin{aligned} &=y+x \frac{d y}{d x}=0 \\ &=x \frac{d y}{d x}=-y \\ &=\frac{d y}{d x}=\frac{-y}{x} \\ &=m_{1}=\frac{d y}{d x}=\frac{-y}{x}---(3) \end{aligned}

Second curve is \begin{aligned} &x^{2}+y^{2}=8\end{aligned}

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

At (2,2) in eq (3) , we get
\begin{aligned} &=m_{1}=\frac{-2}{2}=-1 \\ &=m_{1}=-1 \end{aligned}

At (2, 2) in eq (4) , we get

\begin{aligned} &=m_{1}=\frac{-2}{2}=-1 \\ &=m_{1}=-1 \end{aligned}

Clearly, \left(\frac{d y}{d x}\right)_{c_{1}}=\left(\frac{d y}{d x}\right)_{c_{2}} at (2,2)

At (-2,-2) in eq (3) , we get

\begin{aligned} &=m_{1}=\frac{-(-2)}{-2}=-1 \\ &=m_{1}=-1 \end{aligned}

At (-2,-2) in eq (4) , we get
\begin{aligned} &=m_{1}=\frac{-(-2)}{-2}=-1 \\ &=m_{1}=-1 \end{aligned}

Clearly, \left(\frac{d y}{d x}\right)_{c_{1}}=\left(\frac{d y}{d x}\right)_{c_{2}}  at  (-2,-2)

So, given two curves touch each other at (2,2).

Simillarly, it can be seen that two curves touch each other at (-2, -2)

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