At which point does the tangent to the curve | vedclass

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(D) Given the curve equation: $\sqrt{x} + \sqrt{y} = \sqrt{a}$.Differentiating both sides with respect to $x$:$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y

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$(0, a)$

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(D) Given the curve equation: $\sqrt{x} + \sqrt{y} = \sqrt{a}$.Differentiating both sides with respect to $x$:$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = 0$.Solving for $\frac{dy}{dx}$:$\frac{dy}{dx} = -\sqrt{\frac{y}{x}}$.For the tangent to be perpendicular to the $x$-axis,the slope $\frac{dy}{dx}$ must be undefined,which occurs when the denominator is zero.Thus,$x = 0$.Substituting $x = 0$ into the original equation $\sqrt{x} + \sqrt{y} = \sqrt{a}$:$\sqrt{0} + \sqrt{y} = \sqrt{a} \implies \sqrt{y} = \sqrt{a} \implies y = a$.Therefore,the point is $(0, a)$.

At which point does the tangent to the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ become perpendicular to the $x$-axis?

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