Find the equations of the tangent and the normal to the curve `x=asect ,\ \ y=btant` at `t` at indicated points.

Explanation

`x=a sec t` and `y=b tan t`

`frac{d x}{d t}=a sec t tan t` and `frac{d y}{d t}=b sec ^{2} t`

`therefore frac{d y}{d x}=frac{frac{d y}{d t}}{frac{d x}{d t}}=frac{b sec ^{2} t}{a sec t tan t}=frac{b}{a} cosec t`

Slope of tangent, `m=(frac{d y}{d x})_{t=t}=frac{b}{a} cosec t`

` Now,(x_{1}, y_{1})=(a sec t, b tan t)`

Equation of tangent is ,

`y-y_{1}=m(x-x_{1})`

`Rightarrow y-b tan t=frac{b}{a} cosec t(x-a sec t)`

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