# Find the equation

Question:

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 $(a \cos \theta, b \sin \theta)$

Solution:

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 $)\left(x-x_{1}\right)$

$y-b \sin \theta=-\frac{b^{2}}{\cot \theta a^{2}}(x-a \cos \theta)$