The coordinates of the point on the ellipse $16 x^2+9 y^2=400$ where the ordinate decreases at the same rate at which the abscissa increases, are
Answer & explanation
Correct answer: option 1
We have,
$16 x^2+9 y^2=400$
$\Rightarrow 32 x \frac{d x}{d t}+18 y \frac{d y}{d t}=0$ [Differentiating w.r.t. t]
$\Rightarrow 32 x \frac{d x}{d t}-18 y \frac{d x}{d t}=0$ [∵ $-\frac{dy}{dt} = \frac{dx}{dt}$ (given)]
$\Rightarrow 16 x-9 y=0$
$\Rightarrow y=\frac{16 x}{9}$
∴ $16 x^2+9 y^2=400 \Rightarrow 16 x^2+\frac{256}{9} x^2=400 \Rightarrow x= \pm 3$
Now, $y=\frac{16}{9} x$ and $x= \pm 3 \Rightarrow y= \pm \frac{16}{3}$
Hence, the required points are $(3,16 / 3)$ and $(-3,-16 / 3)$