# Find the equation

Question:

Find the equation of the tangent and the normal to the following curves at the indicated points:

$\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$ at $(\mathrm{a} \sec \theta, \mathrm{b} \tan \theta)$

Solution:

finding the slope of the tangent by differentiating the curve

$\frac{x}{a^{2}}-\frac{y}{b^{2}} \frac{d y}{d x}=0$

$\frac{d y}{d x}=\frac{x b^{2}}{y a^{2}}$

$\mathrm{m}(\operatorname{tangent})$ at $(\mathrm{a} \sec \theta, \mathrm{b} \tan \theta)=\frac{\mathrm{b}}{\mathrm{a} \sin \theta}$

normal is perpendicular to tangent so, $m_{1} m_{2}=-1$

$\mathrm{m}($ normal $)$ at $(\mathrm{a} \sec \theta, \mathrm{b} \tan \theta)=-\frac{\mathrm{a} \sin \theta}{\mathrm{b}}$

equation of tangent is given by $y-y_{1}=m($ tangent $)\left(x-x_{1}\right)$

$\mathrm{y}-\mathrm{b} \tan \theta=\frac{\mathrm{b}}{\mathrm{a} \sin \theta}(\mathrm{x}-\operatorname{asec} \theta)$

equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$

$y-b \tan \theta=-\frac{a \sin \theta}{b}(x-a \sec \theta)$