Find the points on the curve frac{x^{2}}{9}+f | vedclass

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Question

(A) The equation of the given curve is $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$. Differentiating both sides with respect to $x$,we get: $\frac{2x}{9} + \f

Vedclass · Accepted Answer

$(3, 0)$ and $(-3, 0)$

Vedclass · Answer

(A) The equation of the given curve is $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$. Differentiating both sides with respect to $x$,we get: $\frac{2x}{9} + \frac{2y}{16} \cdot \frac{dy}{dx} = 0$. This simplifies to $\frac{dy}{dx} = -\frac{16x}{9y}$. The tangent is parallel to the $y$-axis when the slope $\frac{dy}{dx}$ is undefined,which occurs when the denominator $y = 0$. Substituting $y = 0$ into the equation of the curve: $\frac{x^{2}}{9} + \frac{0^{2}}{16} = 1$ $\Rightarrow x^{2} = 9$ $\Rightarrow x = \pm 3$. Thus,the points are $(3, 0)$ and $(-3, 0)$.

Find the points on the curve $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$ at which the tangents are parallel to the $y$-axis.

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