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

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Question

(C) Given the curve equation: $\frac{x^2}{4} + \frac{y^2}{25} = 1$.Differentiating with respect to $x$:$\frac{2x}{4} + \frac{2y}{25} \frac{dy}{dx} = 0

Vedclass · Accepted Answer

$(0, \pm 5)$

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(C) Given the curve equation: $\frac{x^2}{4} + \frac{y^2}{25} = 1$.Differentiating with respect to $x$:$\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}$.For the tangent to be parallel to the $x$-axis,the slope $\frac{dy}{dx}$ must be $0$.Setting $\frac{dy}{dx} = 0$ implies $x = 0$.Substituting $x = 0$ into the original curve equation:$\frac{0^2}{4} + \frac{y^2}{25} = 1$$\frac{y^2}{25} = 1$$y^2 = 25$$y = \pm 5$.Therefore,the points on the curve where the tangent is 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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