The points on the ellipse 16x^{2} + 9y^{2} = | vedclass

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Question

(A) Given that the ordinate $y$ decreases at the same rate at which the abscissa $x$ increases, we have $\frac{dy}{dt} = -\frac{dx}{dt}$.The equation

Vedclass · Accepted Answer

$\left(3, \frac{16}{3}\right)$ and $\left(-3, -\frac{16}{3}\right)$

Vedclass · Answer

(A) Given that the ordinate $y$ decreases at the same rate at which the abscissa $x$ increases, we have $\frac{dy}{dt} = -\frac{dx}{dt}$.The equation of the ellipse is $16x^{2} + 9y^{2} = 400$.Differentiating both sides with respect to $t$, we get:$16(2x) \frac{dx}{dt} + 9(2y) \frac{dy}{dt} = 0$$32x \frac{dx}{dt} + 18y \frac{dy}{dt} = 0$$16x \frac{dx}{dt} + 9y \frac{dy}{dt} = 0$.Substituting $\frac{dy}{dt} = -\frac{dx}{dt}$ into the equation:$16x \frac{dx}{dt} + 9y \left(-\frac{dx}{dt}ight) = 0$$(16x - 9y) \frac{dx}{dt} = 0$.Since $\frac{dx}{dt} 
eq 0$ for the points to be moving, we have $16x - 9y = 0$, which implies $y = \frac{16}{9}x$.Substituting $y = \frac{16}{9}x$ into the ellipse equation:$16x^{2} + 9\left(\frac{16}{9}xight)^{2} = 400$$16x^{2} + 9 \cdot \frac{256}{81}x^{2} = 400$$16x^{2} + \frac{256}{9}x^{2} = 400$$\frac{144x^{2} + 256x^{2}}{9} = 400$$\frac{400x^{2}}{9} = 400$$x^{2} = 9 \Rightarrow x = \pm 3$.If $x = 3$, then $y = \frac{16}{9}(3) = \frac{16}{3}$.If $x = -3$, then $y = \frac{16}{9}(-3) = -\frac{16}{3}$.Thus, the points are $\left(3, \frac{16}{3}ight)$ and $\left(-3, -\frac{16}{3}ight)$.

The points on the ellipse $16x^{2} + 9y^{2} = 400$ at which the ordinate decreases at the same rate at which the abscissa increases are given by:

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