#### Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 7 Maths Textbook Solution.

ANSWER: Equation of tangent, $\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1$

Equation  of normal , $a x \sec \theta-b y \operatorname{cosec} \theta=\left(a^{2}-b^{2}\right)$

HINTS:

Differentiating the given equation with respect to x.

GIVEN:

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text { at }(a \cos \theta, b \sin \theta)$

Solution:

\begin{aligned} &\Rightarrow \frac{2 x}{a^{2}}+\frac{2 y}{b^{2}} \cdot \frac{d y}{d x}=0 \\ &\Rightarrow \frac{2 y}{b^{2}} \cdot \frac{d y}{d x}=\frac{-2 x}{a^{2}} \\ &\Rightarrow \frac{d y}{d x}=\frac{-x b^{2}}{y a^{2}} \end{aligned}

Slope of tangent ,

\begin{aligned} m &=\left(\frac{d y}{d x}\right)_{\mid a \cos \theta, b \sin \theta]} \\ &=\frac{-a \cos \theta\left(b^{2}\right)}{b \sin \theta\left(a^{2}\right)} \\ &=\frac{-b \cos \theta}{a \sin \theta} \end{aligned}

Equation of tangent is,

\begin{aligned} &y-y_{1}=m\left(x-x_{1}\right) \\ &\Rightarrow y-b \sin \theta=\frac{-b \cos \theta}{a \sin \theta}(x-a \cos \theta) \\ &\Rightarrow a y \sin \theta-a b \sin ^{2} \theta=-b x \cos \theta+a b \cos ^{2} \theta \\ &\Rightarrow b x \cos \theta+a y \sin \theta=a b \end{aligned}

Dividing by ab

$\therefore \frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1$

Equation of Normal  is,

\begin{aligned} &y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right) \\ &y-b \sin \theta=\frac{a \sin \theta}{b \cos \theta}(x-a \cos \theta) \\ &\Rightarrow b y \cos \theta-b^{2} \sin \theta \cos \theta=a x \sin \theta-a^{2} \sin \theta \cos \theta \\ &\Rightarrow a x \sin \theta-b y \cos \theta=\left(a^{2}-b^{2}\right) \sin \theta \cos \theta \end{aligned}

Dividing by $\sin \theta \cos \theta$

$\therefore a x \sec \theta-b y \operatorname{cosec} \theta=\left(a^{2}-b^{2}\right)$