#### Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 8 Maths Textbook Solution.

ANSWER: Equation of tangent, $\frac{x}{a} \sec \theta-\frac{y}{b} \tan \theta=1$

Equation  of normal , $a x \cos \theta+b y \cot \theta=\left(a^{2}+b^{2}\right)$

HINTS:

Differentiating the given curve  with respect to x and find its slope.

GIVEN:

$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \text { at }(\operatorname{asec} \theta, b \tan \theta)$

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 ,

$m=\left(\frac{d y}{d x}\right)_{(a \sec \theta, b \tan \theta)}$

$=\frac{a \sec \theta \cdot b^{2}}{b \tan \theta \cdot a^{2}}$

$=\frac{b}{a} \cdot \frac{1}{\cos \theta} \cdot \frac{1}{\frac{\sin \theta}{\cos \theta}}$

$\left[\because \sec \theta=\frac{1}{\cos \theta} \cdot \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$

$=\frac{b}{a} \cdot \frac{1}{\sin \theta}=\frac{b}{a} \cos e c \theta$

$\therefore m=\left(\frac{d y}{d x}\right)_{(\operatorname{asec} \theta, b \tan \theta)}=\frac{b}{a \sin \theta}$

Equation of tangent is,

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

\begin{aligned} &\Rightarrow a y \sin \theta \cos \theta-a b \sin ^{2} \theta=b x \cos \theta-a b \\ &\Rightarrow a y \sin \theta \cos \theta-b x \cos \theta=a b \sin ^{2} \theta-a b \\ &\Rightarrow a y \sin \theta \cos \theta-b x \cos \theta=-a b\left(1-\sin ^{2} \theta\right) \\ &\Rightarrow a y \sin \theta \cos \theta-b x \cos \theta=-a b \cos ^{2} \theta \quad\left[\because 1-\sin ^{2} \theta=\cos ^{2} \theta\right] \end{aligned}

Divinding by $-a b \cos ^{2} \theta$

\begin{aligned} &\Rightarrow \frac{a y \sin \theta \cos \theta-b x \cos \theta}{-a b \cos ^{2} \theta}=1 \\ &\Rightarrow \frac{a y \sin \theta \cos \theta}{-a b \cos ^{2} \theta}-\frac{b x \cos \theta}{-a b \cos ^{2} \theta}=1 \\ &\Rightarrow \frac{-y}{b} \frac{\sin \theta}{\cos \theta}-\frac{x}{a} \frac{1}{\cos \theta}=1 \\ &\therefore \frac{x}{a} \sec \theta-\frac{y}{b} \tan \theta=1 \end{aligned}

Equation of Normal  is,

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

Dividing by $\tan \Theta$

$\therefore a x \cos \theta+b y \cot \theta=\left(a^{2}+b^{2}\right)$