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 $(a \cos \theta, b \sin \theta)$
finding the slope of the tangent by differentiating the curve
$\frac{x}{a^{2}}+\frac{y}{b^{2}} \frac{d y}{d x}=0$
$\frac{d y}{d x}=-\frac{x a^{2}}{y b^{2}}$
$\mathrm{m}(\operatorname{tangent}) \mathrm{at}(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)=-\frac{\cot \theta \mathrm{a}^{2}}{\mathrm{~b}^{2}}$
normal is perpendicular to tangent so, $m_{1} m_{2}=-1$
$\mathrm{m}($ normal $)$ at $(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)=\frac{\mathrm{b}^{2}}{\cot \theta \mathrm{a}^{2}}$
equation of tangent is given by $y-y_{1}=m($ tangent $)\left(x-x_{1}\right)$
$y-b \sin \theta=-\frac{\cot \theta a^{2}}{b^{2}}(x-a \cos \theta)$
equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$
$y-b \sin \theta=-\frac{b^{2}}{\cot \theta a^{2}}(x-a \cos \theta)$