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# Solution for 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) - CBSE (Science) Class 12 - Mathematics

ConceptRate of Change of Bodies Or Quantities

#### 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)

#### Solution

(a) (3, 16/3)

$\text {According to the question,}$

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

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

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

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

$\Rightarrow 32x = 18y$

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

$\text { Now,}$

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

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

$\Rightarrow 81 y^2 + 144 y^2 = 6400$

$\Rightarrow 225 y^2 = 6400$

$\Rightarrow y^2 = \frac{6400}{225}$

$\Rightarrow y = \sqrt{\frac{6400}{225}}$

$\Rightarrow y = \frac{16}{3} or - \frac{16}{3}$

$\text { So,}$

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

$\text { or }$

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

$\Rightarrow x = 3 or - 3$

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

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Solution 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) Concept: Rate of Change of Bodies Or Quantities.
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