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
$(3, - 3)$

Solution:

We have,

$\begin{array}{l} 16 x^{2} + 9 y^{2} = 400 \\ \Rightarrow 32 x \frac{d x}{d t} + 18 y \frac{d y}{d t} = 0 [D i f f e r e n t i a t i n g w . r . t . t] \end{array}$

$\Rightarrow 32 x \frac{d x}{d t} - 18 y \frac{d x}{d t} = 0$ $[∵ - \frac{d y}{d t} = \frac{d x}{d t} (given)]$

$\begin{array}{l} \Rightarrow 16 x - 9 y = 0 \\ \Rightarrow y = \frac{16 x}{9} \end{array}$

$∴ 16 x^{2} + 9 y^{2} = 400 \Rightarrow 16 x^{2} + \frac{256}{9} x^{2} = 400 \Rightarrow x = \pm 3$

Now $y = \frac{16}{9} x and x = \pm 3 \Rightarrow y = \pm \frac{16}{3}$

Hence , the required points are $(3, 16 / 3)$ and $(- 3, - 16 / 3)$

The coordinates of the point on the ellipse $16 x^{2} + 9 y^{2} = 400$ where the ordinate decreases at the same rate at which the abscissa increases, are

Solution:

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