Question

The coordinates of the point on the ellipse 16x² + 9y² = 400 where the ordinate decreases at the same rate at which the abscissa increases, are
(a) (3, 16/3)
(b) (−3, 16/3)
(c) (3, −16/3)
(d) (3, −3)

Answer 1

(a) (3, 16/3)

$$\begin{gathered} \text{According to the question,} \\ \frac{dy}{dt}=\frac{-dx}{dt} \\ 16x^2+9y^2=400 \\ \Rightarrow 32x\frac{dx}{dt}+18y\frac{dy}{dt}=0 \\ \Rightarrow 32x\frac{dx}{dt}=-18y\frac{dy}{dt} \\ \Rightarrow 32x=18y \\ \Rightarrow x=\frac{9y}{16}...\left(1\right) \\ \mathrm{Now}, \\ 16{\left(\frac{9y}{16}\right)}^2+9y^2=400 \\ \Rightarrow \frac{81y^2}{16}+9y^2=400 \\ \Rightarrow 81y^2+144y^2=6400 \\ \Rightarrow 225y^2=6400 \\ \Rightarrow y^2=\frac{6400}{225} \\ \Rightarrow y=\sqrt{\frac{6400}{225}} \\ \Rightarrow y=\frac{16}{3}\mathrm{or}-\frac{16}{3} \\ \mathrm{So}, \\ x=\frac{9}{16}\times \frac{16}{3}\left[\mathrm{Using}\left(1\right)\right] \\ \mathrm{or} \\ x=-\frac{9}{16}\times \frac{16}{3} \\ \Rightarrow x=3\mathrm{or}-3 \\ \text{So, the required point is}\left(3,\frac{16}{3}\right)\text{.} \end{gathered}$$