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

विकल्प

• (3, 16/3)

•  (−3, 16/3)

•  (3, −16/3)

• (3, −3)

Answer 1

 (3, 16/3)

$$\text{According to the question,}$$

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

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

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

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

$$\Rightarrow 32x=18y$$

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

$$\text{ Now,}$$

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

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

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

$$\Rightarrow 225y^2=6400$$

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

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

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

$$\text{ So,}$$

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

$$\text{ or }$$

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

$$\Rightarrow x=3\text{ or }-3$$

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