Question

The points of the ellipse $16x^2 + 9y^2 = 400$ at which the ordinate decreases at the same rate at which the abscissa increases is/are given by

• A. $\left(3, \frac{16}{3}\right)\&\left(-3, \frac{-16}{3}\right)$
• B. $\left(3, \frac{-16}{3}\right)\&\left(-3, \frac{16}{3}\right)$
• C. $\left(\frac{1}{16}, \frac{1}{9}\right)\& \left(-\frac{1}{16}, -\frac{1}{9}\right)$
• D. $\left(\frac{1}{16}, -\frac{1}{9}\right)\& \left(-\frac{1}{16}, \frac{1}{9}\right)$

Answer 1

Answer: (A)

$\left(3, \frac{16}{3}\right)\&\left(-3, \frac{-16}{3}\right)$

Answer 2

$\frac{x^{2}}{25}+\frac{y^{2}}{\frac{400}{9}}=1$
$\left(5\,cos\theta, \frac{20}{3}sin\theta\right)$
$x=5cos\theta, y=\frac{20}{3}sin\theta$
$\frac{dx}{d\theta}=-5\,sin\,\theta, \frac{dy}{d\theta}=\frac{20}{3}\,cos\,\theta$
$\frac{dx}{d\theta}=-\frac{dy}{d\theta}$
$-5sin\theta=-\frac{20}{3}cos\,\theta$
$tan\theta=4/3 \Rightarrow cos\theta=3/5$ or $-3/5$
$sin\theta=4/5$ or $-4/5$
Points are $\left(3, \frac{16}{3}\right)$ and $\left(-3, \frac{-16}{3}\right)$