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

$(0, 4)$ and $(0, -4)$

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: $\frac{2x}{9} + \frac{2y}{16} \cdot \frac{dy}{dx} = 0$. $\Rightarrow \frac{dy}{dx} = -\frac{16x}{9y}$. The tangent is parallel to the $x$-axis if the slope of the tangent is $0$,i.e.,$\frac{dy}{dx} = 0$. This implies $-\frac{16x}{9y} = 0$,which is possible only if $x = 0$. Substituting $x = 0$ into the equation of the curve: $\frac{0^{2}}{9} + \frac{y^{2}}{16} = 1 \Rightarrow y^{2} = 16 \Rightarrow y = \pm 4$. Thus,the points on the curve where the tangents are parallel to the $x$-axis are $(0, 4)$ and $(0, -4)$.

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

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