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  • Provide Solution For R.D. Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Questions Question 19 Maths Textbook Solution.

Provide Solution For  R.D. Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Questions Question 19 Maths Textbook Solution.

Answers (1)

Answer:

            (c) Is the correct option 

Hint:

        Equation of normal y-y_{1}=m\left ( x-x_{1} \right )

Given:

             x=a \cos ^{3} \theta, y=a \sin ^{3} \theta

Solution:

x=a \cos ^{3} \theta            (1)

Differentiating w.r.t \theta

\begin{aligned} &\frac{d x}{d \theta}=a \cdot 3 \cos ^{2} \theta(-\sin \theta) \\ &=-3 a \cos ^{2} \theta \sin \theta \\ &\left.\frac{d x}{d \theta}\right|_{\theta-\frac{\pi}{4}}=-3 a \times \frac{1}{2} \times \frac{1}{\sqrt{2}}=-\frac{3 a}{2 \sqrt{2}} \\ &y=a \sin ^{3} \theta \end{aligned}

Differentiating w.r.t \theta

\begin{aligned} &\frac{d y}{d \theta}=a \times 3 \sin ^{2} \theta \cos \theta \\ &\left.\frac{d y}{d \theta}\right|_{\theta-\frac{\pi}{4}}=3 a \times \frac{1}{2} \times \frac{1}{\sqrt{2}}=\frac{3 a}{2 \sqrt{2}} \end{aligned}

Slope of normal of the given curve M(N)=-\frac{d x}{d y}=-\left(\frac{\frac{d x}{d \theta}}{\frac{d y}{d \theta}}\right)=-\left(\frac{\frac{3 \alpha}{2 \sqrt{2}}}{\frac{-3 \alpha}{2 \sqrt{2}}}\right)=1

Equation of normal at \theta =\frac{\pi }{4}

\begin{aligned} &y-\frac{a}{(\sqrt{2})^{3}}=1\left(x-\left(\frac{a}{\sqrt{2}}\right)^{3}\right) \\ &y=x \end{aligned}

 

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