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. $\left(\begin{array}{c}3,\frac{16}{3}\end{array}\right)$ and $\left(\begin{array}{c}-3,\frac{-16}{3}\end{array}\right)$
• B. $\left(\begin{array}{c}3,\frac{-16}{3}\end{array}\right)$ and $\left(\begin{array}{c}-3,\frac{16}{3}\end{array}\right)$
• C. $\left(\begin{array}{c}\frac{1}{16},\frac{1}{9}\end{array}\right)$ and $\left(\begin{array}{c}-\frac{1}{16},-\frac{1}{9}\end{array}\right)$
• D. $\left(\begin{array}{c}\frac{1}{16},-\frac{1}{9}\end{array}\right)$ and $\left(\begin{array}{c}-\frac{1}{16},\frac{1}{9}\end{array}\right)$

Answer 1

The correct option is A $\left(\begin{array}{c}3,\frac{16}{3}\end{array}\right)$ and $\left(\begin{array}{c}-3,\frac{-16}{3}\end{array}\right)$

$16x^2+9y^2=400$

Differentiating w.r.t time, we get

$32x\frac{dx}{dt}+18y\frac{dy}{dt}=0$

Given that the ordinate decrease at the same rate as the abscissa increases, $\frac{dy}{dt}=\frac{-dx}{dt}$

$\Rightarrow 32x=18y$ or $16x=9y$

Substituting $x=\frac{9y}{16}$ in the equation, we get

$16\times \frac{9y}{16}\times \frac{9y}{16}+9y^2=400$

$\Rightarrow y^2(\frac{81}{16}+9)=400$

$\Rightarrow y^2=\frac{400\times 16}{81+9\times 16}$

$\therefore y^2=\frac{400\times 16}{225}=\frac{256}{9}$

$\therefore y=\pm \frac{16}{3}$

Correspondingly, $x=\pm 3$

$\therefore$ the points are $(3,\frac{16}{3}),(-3,\frac{-16}{3})$