Find the equation of the tangent and the normal to the following curves at the indicated points:
$x=a \sec t, y=b \tan t a t t$
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 $)\left(x-x_{1}\right)$
$y-b \tan t=-\frac{\operatorname{asin} t}{b}(x-\operatorname{asec} t)$
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