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

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Question

(A) The given 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}

Vedclass · Accepted Answer

$(2,0)$ and $(-2,0)$

Vedclass · Answer

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

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

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