Question

Find area of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$.

Answer 1

We know $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is an ellipse,

Let us consider general point by $(x,y)=(acos\theta ,bsin\theta )$

So,

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

$\begin{array}{c}x=5\cos \theta ,y=4sin\theta \\ \frac{dx}{d\theta }=-5\cos \theta ,\frac{dy}{d\theta }=4\cos \theta \\ Slopeof\tan gent=\frac{dy}{dx}\\ =\frac{\frac{dt}{d\theta }}{\frac{dx}{d\theta }}=\frac{4\cos \theta }{-5\sin \theta }=\frac{-4\cot \theta }{5}\\ Slopeofnormal=\frac{-1}{\frac{dy}{dx}}=\frac{-1}{\frac{-4\cot \theta }{5}}=\frac{5}{4\cot \theta }=\frac{5\tan \theta }{4}\end{array}$

Now, normal equation

$\begin{array}{c}(5\cos \theta ,4\sin \theta )=\\ y-4\sin \theta =\frac{-1}{\frac{dy}{dx}}(x-5\cos \theta )\\ y-4\sin \theta =\frac{5\tan \theta }{4}(x-5\cos \theta )\\ 4y-16\sin \theta =5x\tan \theta -25\sin \theta \\ 4y-16\sin \theta -5x\tan \theta +25\sin \theta =0\end{array}$

Line distance from ellipse's centre

$\begin{array}{c}d=\frac{(5^2-4^2)\sin \theta }{{(1+{\tan }^2\theta )}^{\frac{1}{2}}}\\ =\frac{9\sin \theta }{\sec \theta }\\ =9\sin \theta \cos \theta \\ =\frac{9\sin 2\theta }{2}\end{array}$

$sin2\theta$ maximum value $=$ 1

Maximum value $d_{max}=\frac{9}{2}$