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Explain solution for RD Sharma maths Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Question, Question 31 maths textbook solution.

Answers (1)

Answer : (b) is the correct option

Hint : Multiply the slope of both curve

Given : x^{3}-3 x y^{2}+2=0 \text { And } 3 x^{2} y-y^{3}=2

Solution :

Given x^{3}-3 x y^{2}+2=0                                             (1)

Differentiating (1) we get

\begin{aligned} &3 x^{2}-3 y^{2}-3 x(2 y) \frac{d y}{d x}=0 \\ &3\left(x^{2}-y^{2}\right)=6 x y m_{1} \\ &m_{1}=\frac{x^{2}-y^{2}}{2 x y} \end{aligned}

And  3 x^{2} y-y^{3}-2=0                                                (2)

Differentiating (2), we get

\begin{aligned} &6 x y-3 x^{2} \frac{d y}{d x}-3 y^{2} \frac{d y}{d x}=0 \\ &6 x y=6 m_{2}\left(y^{2}-x^{2}\right) \\ &m_{2}=\frac{2 x y}{y^{2}-x^{2}} \end{aligned}

m_{1} \times m_{2}=\frac{x^{2}-y}{2 x y} \times \frac{2 x y}{y^{2}-x^{2}}=-1

Equation (1) and (2) at right angle

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