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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.

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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