`
Find the equation of the tangent and the normal to the following curves at the indicated points:
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at $(\sqrt{2} a, b)$
finding slope of the tangent by differentiating the curve
$\frac{x}{a^{2}}-\frac{y}{b^{2}} \frac{d y}{d x}=0$
$\frac{d y}{d x}=\frac{x b^{2}}{y a^{2}}$
$\mathrm{m}(\operatorname{tangent})$ at $(\sqrt{2} \mathrm{a}, \mathrm{b})=\frac{\sqrt{2} \mathrm{ab}^{2}}{\mathrm{ba}^{2}}$
normal is perpendicular to tangent so, $m_{1} m_{2}=-1$
$\mathrm{m}$ (normal) at $(\sqrt{2} \mathrm{a}, \mathrm{b})=-\frac{\mathrm{ba}^{2}}{\sqrt{2} \mathrm{ab}^{2}}$
equation of tangent is given by $y-y_{1}=m(\operatorname{tangent})\left(x-x_{1}\right)$
$y-b=\frac{\sqrt{2} a b^{2}}{b a^{2}}(x-\sqrt{2} a)$
equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$
$y-b=-\frac{b a^{2}}{\sqrt{2} a b^{2}}(x-\sqrt{2} a)$
.jpg)