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

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

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$(0, \pm 5)$

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(C) Given the equation of the curve is $\frac{x^2}{4} + \frac{y^2}{25} = 1$.Differentiating both sides with respect to $x$,we get:$\frac{2x}{4} + \frac{2y}{25} \frac{dy}{dx} = 0$Simplifying the expression:$\frac{x}{2} + \frac{2y}{25} \frac{dy}{dx} = 0$$\frac{2y}{25} \frac{dy}{dx} = -\frac{x}{2}$$\frac{dy}{dx} = -\frac{25x}{4y}$For the tangent to be parallel to the $x$-axis,the slope of the tangent must be $0$,i.e.,$\frac{dy}{dx} = 0$.Setting the derivative to zero:$-\frac{25x}{4y} = 0 \implies x = 0$.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 on the curve where 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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