If the tangent to the curve frac{x^2}{a^2} + | vedclass

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(C) Differentiating the given equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with respect to $x$,we get:$\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{d

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$(0, \pm b)$

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(C) Differentiating the given equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with respect to $x$,we get:$\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0$Solving for $\frac{dy}{dx}$,we get:$\frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$If the tangent is parallel to the $x$-axis,its slope must be zero:$\frac{dy}{dx} = 0 \implies -\frac{b^2 x}{a^2 y} = 0 \implies x = 0$Substituting $x = 0$ into the original equation of the ellipse:$\frac{0^2}{a^2} + \frac{y^2}{b^2} = 1 \implies y^2 = b^2 \implies y = \pm b$Therefore,the points of tangency are $(0, \pm b)$.

If the tangent to the curve $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is parallel to the $x$-axis,then what is the point of tangency?

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