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

पर्याय

• (3, 16/3)

•  (−3, 16/3)

•  (3, −16/3)

• (3, −3)

Answer 1

 (3, 16/3)

\[\text {According to the question,}\]

\[\frac{dy}{dt} = \frac{- dx}{dt}\]

\[16 x^2 + 9 y^2 = 400\]

\[ \Rightarrow 32x\frac{dx}{dt} + 18y\frac{dy}{dt} = 0\]

\[ \Rightarrow 32x\frac{dx}{dt} = - 18y\frac{dy}{dt}\]

\[ \Rightarrow 32x = 18y\]

\[ \Rightarrow x = \frac{9y}{16} . . . \left( 1 \right)\]

\[\text { Now,} \]

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

\[ \Rightarrow \frac{81 y^2}{16} + 9 y^2 = 400\]

\[ \Rightarrow 81 y^2 + 144 y^2 = 6400\]

\[ \Rightarrow 225 y^2 = 6400\]

\[ \Rightarrow y^2 = \frac{6400}{225}\]

\[ \Rightarrow y = \sqrt{\frac{6400}{225}}\]

\[ \Rightarrow y = \frac{16}{3} or - \frac{16}{3}\]

\[\text { So,} \]

\[x = \frac{9}{16} \times \frac{16}{3} \left[ \text { Using } \left( 1 \right) \right] \]

\[\text { or }\]

\[x = - \frac{9}{16} \times \frac{16}{3}\]

\[ \Rightarrow x = 3 \text { or } - 3\]

\[\text { So, the required point is }\left( 3, \frac{16}{3} \right).\]