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

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(3, $\frac{16}{3}$) \text {According to the question,} dy/dt = - dx/dt 16 x^2 + 9 y^2 = 400 \Rightarrow 32xdx/dt + 18ydy/dt = 0 \Rightarrow 32xdx/dt = - 18ydy/dt \Rightarrow 32x = 18y \Rightarrow x = 9y/16 . . . \left( 1 \right) \text { Now,} 16 \left( 9y/16 \right)^2 + 9 y^2 = 400 \Rightarrow 81 y^$\frac{2}{16}$ + 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).

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