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

Answer 1

The correct option is A $(3,\frac{16}{3})\&(-3,\frac{-16}{3})$

Standard equation of ellipse is,$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Consider the equation of ellipse is
$\frac{x^2}{25}+\frac{y^2}{\frac{400}{9}}=1$

Here,$a=25$ and $b=\frac{20}{3}$

The parametric form of the ellipse,
$x=a\cos \theta =5\cos \theta$ and $y=b\sin \theta =\frac{20}{3}\sin \theta$
$\therefore \frac{dx}{d\theta }=-5\sin \theta$ and $\frac{dy}{d\theta }=\frac{20}{3}\cos \theta$
As per the question
$\frac{dx}{d\theta }=-\frac{dy}{d\theta }$
$\Rightarrow -5\sin \theta =-\frac{20}{3}\cos \theta$
$\Rightarrow \tan \theta =\frac{4}{3}$
$\cos \theta =\pm \frac{3}{5}\sin \theta =\pm \frac{4}{5}$
Hence the coordinates will be,
$(5\cos \theta ,\frac{20}{3}\sin \theta )$
$=(3,\frac{16}{3})\&(-3,-\frac{16}{3})$