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 $(a cos θ, b sin θ)$ .

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Solution

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

Differentiating both sides w.r.t. $x,$

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

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

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

Slope of tangent, $m = {(\frac{d y}{d x})}_{(a cos θ, b sin θ)} = \frac{- a cos θ (b^{2})}{b sin θ (a^{2})}$

$= \frac{- b cos θ}{a sin θ}$

$(x_{1}, y_{1}) = (a cos θ, b sin θ)$

Equation of tangent :
$y - y_{1} = m (x - x_{1})$

$\Rightarrow$ $y - b sin θ = \frac{- b cos θ}{a sin θ} (x - a cos θ)$

$\Rightarrow$ $a y sin θ - a b {sin}^{2} θ = - b x cos θ + a b {cos}^{2} θ$

$\Rightarrow$ $b x cos θ + a y sin θ = a b$

Dividing by $a b,$

$\Rightarrow$ $\frac{x}{a} cos θ + \frac{y}{b} sin θ = 1$

Equation of normal:
$y - y_{1} = \frac{- 1}{m} (x - x_{1})$

$\Rightarrow$ $y - b sin θ = \frac{a sin θ}{b cos θ} (x - a cos θ)$

$\Rightarrow$ $b y cos θ - b^{2} sin θ cos θ = a x sin θ - a^{2} sin θ cos θ$

$\Rightarrow$ $a x sin θ - b y cos θ = (a^{2} - b^{2}) sin θ cos θ$

Dividing by $sin θ cos θ$

$\Rightarrow$ $a x sec θ - b y c o s e c θ = (a^{2} - b^{2})$

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