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

(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}