Question

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 $(a\cos \theta ,b\sin \theta )$.

Answer 1

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

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

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

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

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

Slope of tangent, $m={\left(\frac{dy}{dx}\right)}_{(a\cos \theta ,b\sin \theta )}=\frac{-a\cos \theta \left(b^2\right)}{b\sin \theta \left(a^2\right)}$

$=\frac{-b\cos \theta }{a\sin \theta }$

$(x_1,y_1)=(a\cos \theta ,b\sin \theta )$

Equation of tangent :

$y-y_1=m(x-x_1)$

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

$\Rightarrow$ $ay\sin \theta -ab{\sin }^2\theta =-bx\cos \theta +ab{\cos }^2\theta$

$\Rightarrow$ $bx\cos \theta +ay\sin \theta =ab$

Dividing by $ab,$

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

Equation of normal:

$y-y_1=\frac{-1}{m}(x-x_1)$

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

$\Rightarrow$ $by\cos \theta -b^2\sin \theta \cos \theta =ax\sin \theta -a^2\sin \theta \cos \theta$

$\Rightarrow$ $ax\sin \theta -by\cos \theta =(a^2-b^2)\sin \theta \cos \theta$

Dividing by $\sin \theta \cos \theta$

$\Rightarrow$ $ax\sec \theta -bycosec\theta =(a^2-b^2)$