Question

Find the equation of the tangent and the normal to the following curve at the indicated point.
$4x^2+9y^2=36$ at $(3\cos \theta ,2\sin \theta )$.

Answer 1

The given equation is,

$4x^2+9y^2=36$

or

$y^2=\frac{(36-4x^2)}{9}$

Differentiate to get slope of tangent,

$2y\frac{dy}{dx}=-\frac{4}{9}2x$

$\Rightarrow \frac{dy}{dx}=-\frac{4x}{9y}$

Thus the slope of tangent at $(3cos\theta ,2sin\theta )$,

$\frac{dy}{dx}=-\frac{4\cdot 3cos\theta }{9\cdot 2sin\theta }=-\frac{2cot\theta }{3}$

Apply point-slope form to get the equation of tangent.

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

$y=-\frac{2}{3}cot\theta \cdot x+2\frac{{cos}^2\theta }{sin\theta }+2sin\theta$

$y=-\frac{2}{3}cot\theta \cdot x+2cosec\theta$

For finding the equation of the normal.

$m_{normal}=-\frac{1}{m_{tangent}}=\frac{3}{2}tan\theta$

Hence,

Apply point-slope form to get the equation of normal.

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

$y=\frac{3}{2}tan\theta \cdot x-\frac{9}{2}sin\theta +2sin\theta$

$y=\frac{3}{2}tan\theta \cdot x-\frac{5}{2}sin\theta$

(answer)