What is the sum of the intercepts on the coor | vedclass

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(D) Given the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$.Differentiating with respect to $x$,we get $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{d

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(D) Given the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$.Differentiating with respect to $x$,we get $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0$.Thus,$\left( \frac{dy}{dx} \right)_{(x_1, y_1)} = -\sqrt{\frac{y_1}{x_1}}$.The equation of the tangent at $(x_1, y_1)$ is $y - y_1 = -\sqrt{\frac{y_1}{x_1}}(x - x_1)$.For the $x$-intercept,set $y = 0$: $-y_1 = -\sqrt{\frac{y_1}{x_1}}(x - x_1) \implies x - x_1 = y_1 \sqrt{\frac{x_1}{y_1}} = \sqrt{x_1 y_1}$.So,$x = x_1 + \sqrt{x_1 y_1} = \sqrt{x_1}(\sqrt{x_1} + \sqrt{y_1}) = \sqrt{x_1} \sqrt{a} = \sqrt{ax_1}$.For the $y$-intercept,set $x = 0$: $y - y_1 = -\sqrt{\frac{y_1}{x_1}}(-x_1) = \sqrt{x_1 y_1}$.So,$y = y_1 + \sqrt{x_1 y_1} = \sqrt{y_1}(\sqrt{y_1} + \sqrt{x_1}) = \sqrt{y_1} \sqrt{a} = \sqrt{ay_1}$.The sum of the intercepts is $\sqrt{ax_1} + \sqrt{ay_1} = \sqrt{a}(\sqrt{x_1} + \sqrt{y_1}) = \sqrt{a} \cdot \sqrt{a} = a$.

What is the sum of the intercepts on the coordinate axes of the tangent to the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ at the point $(x_1, y_1)$?

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