#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 13 Maths Textbook Solution.

ANSWER: Equation of tangent, $\frac{x x_{0}}{a^{2}}-\frac{y y_{0}}{b^{2}}=1$

Equation  of normal , $y-y_{0}=\frac{\alpha^{2}}{b^{2}} \cdot \frac{y_{0}}{x_{0}}\left(x-x_{0}\right)$

HINTS:

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

GIVEN:

$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \text { at }\left(x_{0}, y_{0}\right)$                        ..............(i)

SOLUTION:

Since $P\left(x_{0}, y_{0}\right)$ lies on the curve(i)

\begin{aligned} &\frac{2 x}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \\ &\frac{2 x}{a^{2}}-\frac{2 y}{b^{2}} \cdot \frac{d y}{d x}=0 \\ &\Rightarrow \frac{d y}{d x}=\frac{x b^{2}}{y a^{2}} \end{aligned}

Slope of tangent,

$\left.\Rightarrow \frac{d y}{d x}\right|_{\left.i x_{0}, y_{0}\right]}=\frac{x_{0} b^{2}}{y_{0} a^{2}}=m$

Equation of tangent is

\begin{aligned} &y-y_{0}=m\left(x-x_{0}\right) \\ &\Rightarrow y-y_{0}=\frac{x_{0} b^{2}}{y_{0} a^{2}}\left(x-x_{0}\right) \end{aligned}

$\Rightarrow \frac{y_{0}}{b^{2}}(y-y_{0})=\frac{x^{0}}{a^2}(x-x_{0})$

\begin{aligned} &\Rightarrow \frac{y y_0}{b^{2}}-\frac{y_0^{2}}{b^{2}}=\frac{x_0 x}{a^{2}}-\frac{x_{0}^{2}}{a^{2}} \\ &\Rightarrow \frac{y y_0}{b^{2}}-\frac{x x_0}{a^{2}}=\frac{y_0^2}{b^{2}}-\frac{x_{0}^{2}}{a^{2}} \end{aligned}

$\Rightarrow \frac{x x_{0}}{a^{2}}-\frac{y y_{o}}{b^{2}}=1$

Again, slope of normal,

\begin{aligned} &m_{1} m_{2}=-1 \\ &\frac{b^{2}}{a^{2}} \cdot \frac{x_{0}}{y_{0}} \cdot m_{2}=-1 \\ &m_{2}=-\frac{a^{2}}{b^{2}} \frac{y^{0}}{x^{0}} \quad a t\left(x_{0}, y_{0}\right) \end{aligned}

Equation of Normal  is ,

$\therefore \mathrm{y}-\mathrm{y}_{0}=-\frac{a^{2}}{b^{2}} \frac{y^{0}}{x^{0}}\left(x-x_{0}\right)$

$\Rightarrow \frac{b^{2}\left(y-y_{0}\right)}{y_{0}}=-\frac{a^{2}\left(x-x_{0}\right)}{x_{0}}$

$\Rightarrow \frac{b^{2} y}{y_{0}}-b^{2}=-\frac{a^{2} x}{x_{0}}+a^{2}$

$\Rightarrow \frac{a^{2} x}{x_{0}}+\frac{b^{2} y}{y_{0}}=a^{2}+b^{2}$