Question

At what points of the ellipse $16x^2+9y^2=400$, does the ordiante decreases at the same rate at which abscissa increases.

Answer 1

$16x^2+9y^2=400--\left(1\right)$

$32x\frac{dx}{dt}+18y\frac{dy}{dt}=0\Rightarrow 9y\frac{dy}{dt}=-16\frac{dx}{dt}---\left(2\right)$

$\frac{dy}{d}=-\frac{dx}{dt}$

$-dy\frac{dy}{dt}=-\frac{}{1}6xdxdx\Rightarrow 9y=16x\Rightarrow y=\frac{16}{9}x$

$y=\frac{16}{9}x\Rightarrow 16x^2+9\frac{256}{81}{x}^2=480$

$x^2+\frac{16}{9}{x}^2=25\Rightarrow \frac{25x^2}{9}=25$

$\Rightarrow x^2-9\Rightarrow x=-3,3$

$y=\frac{16}{9}(-3),\frac{16}{9}\left(3\right)=\frac{-16}{3},\frac{16}{3}$

required $pt$ are $(-3,\frac{-16}{3}),(3,\frac{16}{3})$