Question

Find length of normal to $x=\sqrt{2}\cos t,y=3\sin t$ at $t=\frac{\pi }{4}$.

Answer 1

Given,

$x=\sqrt{2}\cos t$ and $y=3\sin t$

Now,

$\frac{dx}{dt}=-\sqrt{2}\sin t$......(1) and $\frac{dy}{dt}=3\cos t$......(2).

Then $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=-\frac{3\cot t}{\sqrt{2}}$.

Now, ${\frac{dy}{dx}|}_{t=\frac{\pi }{4}}=-\frac{3}{\sqrt{2}}$.

So, the length of the normal to $x=\sqrt{2}\cos t,y=3\sin t$ at $t=\frac{\pi }{4}$ is ${y\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}|}_{x=\frac{\pi }{4}}=\frac{3}{\sqrt{2}}.\frac{\sqrt{11}}{\sqrt{2}}=\frac{3\sqrt{11}}{2}$.