#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 17 Maths Textbook Solution.

ANSWER: Equation of tangent, $2 x \cos \theta+3 y \sin \theta=6$

Equation  of normal , $3 x \sin \theta-2 y \cos \theta-5 \sin \theta \cos \theta=0$

HINTS:

Differentiating the given curve with respect to x

GIVEN:

$4 x^{2}+9 y^{2}=36 a t[3 \cos \theta, 2 \sin \theta]$

SOLUTION:

\begin{aligned} &8 x+18 y \frac{d y}{d x}=0 \\ &\Rightarrow 18 y \frac{d y}{d x}=-8 x \\ &\Rightarrow \frac{d y}{d x}=\frac{-8 x}{18 y}=\frac{-4 x}{9 y} \end{aligned}

Slope of tangent ,

$m=\left(\frac{d y}{d x}\right)_{(3 \cos \theta, 2 \sin \theta)}$

Equation of tangent is,

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-2 \sin \theta=\frac{-2 \cos \theta}{3 \sin \theta}(x-3 \cos \theta)$

\begin{aligned} &\Rightarrow 3 y \sin \theta-6 \sin ^{2} \theta=-2 x \cos \theta+6 \cos ^{2} \theta \\ &\Rightarrow 2 x \cos \theta+3 y \sin \theta=6\left(\cos ^{2} \theta+\sin ^{2} \theta\right) \end{aligned}

$\therefore 2 x \cos \theta+3 y \sin \theta=6$

Equation of Normal  is,

$y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right)$

$\Rightarrow y-2 \sin \theta=\frac{3 \sin \theta}{2 \cos \theta}(x-3 \cos \theta)$

\begin{aligned} &\Rightarrow 2 y \cos \theta-4 \sin \theta \cos \theta=3 x \sin \theta-9 \sin \theta \cos \theta \\ &\Rightarrow 3 x \sin \theta-2 y \cos \theta-5 \sin \theta \cos \theta=0 \end{aligned}