Find the equation

finding slope of the tangent by differentiating $x$ and $y$ with respect to $t$

$\frac{\mathrm{dx}}{\mathrm{dt}}=\operatorname{asecttan} \mathrm{t}$

$\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{bsec}^{2} \mathrm{t}$

Now dividing $\frac{d y}{d t}$ and $\frac{d x}{d t}$ to obtain the slope of tangent

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{b} \operatorname{cosec} \mathrm{t}}{\mathrm{a}}$

$\mathrm{m}(\mathrm{tangent})$ at $\mathrm{t}=\frac{\mathrm{bcosect}}{\mathrm{a}}$

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

$\mathrm{m}$ (normal) at $\mathrm{t}=-\frac{\mathrm{a}}{\mathrm{b}} \sin \mathrm{t}$

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

$\mathrm{y}-\mathrm{btan} \mathrm{t}=\frac{\mathrm{b} \operatorname{cosec} \mathrm{t}}{\mathrm{a}}(\mathrm{x}-\mathrm{asec} \mathrm{t})$

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

$y-b \tan t=-\frac{\operatorname{asin} t}{b}(x-\operatorname{asec} t)$