Find the points on the curve frac{x^{2}}{4} + | vedclass

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(A) Given the equation of the curve: $\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1$. Differentiating both sides with respect to $x$,we get: $\frac{2x}{4} +

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$(0, 5)$ and $(0, -5)$

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(A) Given the equation of the curve: $\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1$. Differentiating both sides with respect to $x$,we get: $\frac{2x}{4} + \frac{2y}{25} \frac{dy}{dx} = 0$ $\frac{x}{2} + \frac{2y}{25} \frac{dy}{dx} = 0$ $\frac{dy}{dx} = -\frac{25x}{4y}$. The tangent is parallel to the $x$-axis if the slope $\frac{dy}{dx} = 0$. Setting the slope to zero: $-\frac{25x}{4y} = 0$,which implies $x = 0$. Substituting $x = 0$ into the original equation of the curve: $\frac{0^{2}}{4} + \frac{y^{2}}{25} = 1$ $\frac{y^{2}}{25} = 1$ $y^{2} = 25$ $y = \pm 5$. Thus,the points at which the tangents are parallel to the $x$-axis are $(0, 5)$ and $(0, -5)$.

Find the points on the curve $\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1$ at which the tangents are parallel to the $x$-axis.

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