Answer
$ \left(\mp\frac{9}{5}, \pm\frac{16}{5}\right)$
Work Step by Step
$16x^2+9y^2=144$ ................... eq (1)
Taking derivative with respect to t
$\frac{d(16x^2+9y^2) }{dt}=\frac{d(144)}{dt}$
$\frac{d(16x^2+9y^2) }{dx} \frac{dx}{dt}=0$
$\frac{d(16x^2+9y^2) }{dx} \frac{dx}{dt} +\frac{d(16x^2+9y^2) }{dy} \frac{dy}{dt}==0$
$32x \frac{dx}{dt} +18y \frac{dy}{dt}=0$
$32x \frac{dx}{dt} =-18y \frac{dy}{dt}$ ................ eq (2)
Since $\frac{dx}{dt}=\frac{dy}{dt}$ ............................ eq (3)
From equation (2) and equation (3)
$\frac{x}{y}=-\frac{18}{32} =-\frac{9}{16} $
$x =-\frac{9}{16} y $ .................. eq (4)
Putting equation (4) in equation (1)
$16 ( -\frac{9}{16}y)^2+9y^2=144$
$ 16(\frac{81y^2}{256})+9y^2=144$
$ \frac{225}{16}y^2=144$
$ y^2=144\times \frac{16}{225}=\frac{256}{25} $
$y=\pm\frac{16}{5}$ ,,,,,,,,,,,,,,,,,, eq (5)
Putting equation (5) in equation (4)
$x=-\frac{9}{16}\times \pm\frac{16}{5}=\mp \frac{9}{5}$
