Find the equation
Question:

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

$4 x^{2}+9 y^{2}=36$ at $(3 \cos \theta, 2 \sin \theta)$

Solution:

finding the slope of the tangent by differentiating the curve

$8 \mathrm{x}+18 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=0$

$\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{4 \mathrm{x}}{9 \mathrm{y}}$
$\mathrm{m}$ (tangent) at $(3 \cos \theta, 2 \sin \theta)=-\frac{2 \cos \theta}{3 \sin \theta}$

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

$\mathrm{m}($ normal $)$ at $(3 \cos \theta, 2 \sin \theta)=\frac{3 \sin \theta}{2 \cos \theta}$

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

$y-2 \sin \theta=-\frac{2 \cos \theta}{3 \sin \theta}(x-3 \cos \theta)$

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

$y-2 \sin \theta=\frac{3 \sin \theta}{2 \cos \theta}(x-3 \cos \theta)$

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