Question

The point on the ellipse $16x^2+9y^2=400$,at which the ordinate decreases at the same rate at which the abscissa increases is

• A. $(3,\frac{16}{3})$
• B. $(-3,\frac{16}{3})$
• C. $(3,\frac{-16}{3})$
• D. $(-4,\frac{-16}{3})$

Answer 1

Answer: (A)

$(3,\frac{16}{3})$

Decrease in ordinate=increase in Abscissa

Thus

$-dy=dx$

$\frac{dy}{dx}=-1$

We have

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

Differentiating woth respect to $x$

$16\times 2x+9\times 2y\frac{dy}{dx}=0$

$32x-18y=0$

$x=\frac{18}{32}y=\frac{9}{16}y$

Putting this in the equation of ellipse

$16\times {\left(\frac{9}{16}y\right)}^2+9y^2=400$

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

$\frac{81y^2+144y^2}{16}=400$

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

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

Thus $x=\frac{9}{16}\times \frac{16}{3}=3$

Thus the point is $(3,\frac{16}{3})$