Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 5 Sub Question 6  Maths Textbook Solution.

ANSWER: Equation of tangent,$y-3 \sin \theta+\sin ^{3} \theta=-\tan ^{3} \theta\left(x-3 \cos \theta+3 \cos ^{3} \theta\right)$

Equation  of normal , $y-3 \sin \theta+\sin ^{3} \theta=\cot ^{3} \theta\left(x -3 \cos \theta+3 \cos ^{3} \theta\right)$

HINTS:

Differentiating  with respect to $\theta$ to get its slope.

GIVEN:

$x=3\cos \theta -\cos ^{2}\theta ,y=3\sin \theta -\sin ^{3}\theta$

SOLUTION:

Upon differentiation

\begin{aligned} &\frac{d x}{d \theta}=3 \cos \theta+3 \cos ^{2} \theta \sin \theta, \frac{d x}{d \theta}=3 \cos \theta-3 \sin ^{2} \theta \cos \theta \\ &\therefore \frac{d y}{d x}=\frac{3 \cos \theta-3 \sin ^{2} \theta \cos \theta}{-3 \sin \theta+3 \cos ^{2} \sin \theta}=-\tan ^{3} \theta \\ &\mathrm{m}(\text { tangent }) \text { at } \theta \text { is }-\tan ^{3} \theta \end{aligned}

The normal is perpendicular to tangent, therefore , $m_{1},m_{2}=-1$

m(normal0 at $\theta$ is $\cot ^{3}\theta$

The equation of tangent is given by,

\begin{aligned} &y-y_{1}=m(\text { tangent })\left(x-x_{1}\right) \\ &\Rightarrow y-3 \sin \theta+\sin ^{3} \theta=-\tan ^{3} \theta\left(x-3 \cos \theta+3 \cos ^{3} \theta\right) \end{aligned}

The equation of Normal  is given by  ,

\begin{aligned} &y-y_{1}=m(\text { normal })\left(x-x_{1}\right) \\ &\Rightarrow y-3 \sin \theta+\sin ^{3} \theta=\cot ^{3} \theta\left(x-3 \cos \theta+3 \cos ^{3} \theta\right) \end{aligned}