The coordinates of the point on the ellipse $16x^2+9y^2=400$ where the ordinate decreases at the same rate at which the abscissa increases, are

Solution:

We have, 

$\begin{array}{l} 16x^2+9y^2=400\\ \Rightarrow 32x\frac{dx}{dt}+18y\frac{dy}{dt}=0 [Differentiating w.r.t. t] \end{array}$

$\Rightarrow 32x\frac{dx}{dt}-18y\frac{dx}{dt}=0$                                 $\left[\because -\frac{dy}{dt}=\frac{dx}{dt}\text{ (given) }\right]$

$\begin{array}{l}\Rightarrow 16x-9y=0\\ \Rightarrow y=\frac{16x}{9}\end{array}$

$\therefore 16x^2+9y^2=400\Rightarrow 16x^2+\frac{256}{9}{x}^2=400\Rightarrow x=\pm 3$

Now $y=\frac{16}{9}x\text{ and }x=\pm 3\Rightarrow y=\pm \frac{16}{3}$

Hence , the required points are $(3 ,16/3)$ and $(-3 ,-16/3)$