The sum of the intercepts on the axes of the | vedclass

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

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$a$

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(A) Given the curve equation: $\sqrt{x} + \sqrt{y} = \sqrt{a}$.Differentiating with respect to $x$: $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0$.$\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}$.The equation of the tangent at point $(x_1, y_1)$ is $Y - y_1 = -\frac{\sqrt{y_1}}{\sqrt{x_1}}(X - x_1)$.$Y\sqrt{x_1} - y_1\sqrt{x_1} = -X\sqrt{y_1} + x_1\sqrt{y_1}$.$X\sqrt{y_1} + Y\sqrt{x_1} = x_1\sqrt{y_1} + y_1\sqrt{x_1} = \sqrt{x_1 y_1}(\sqrt{x_1} + \sqrt{y_1})$.Since $\sqrt{x_1} + \sqrt{y_1} = \sqrt{a}$,we have $X\sqrt{y_1} + Y\sqrt{x_1} = \sqrt{a}\sqrt{x_1 y_1}$.Dividing by $\sqrt{a}\sqrt{x_1 y_1}$: $\frac{X}{\sqrt{a}\sqrt{x_1}} + \frac{Y}{\sqrt{a}\sqrt{y_1}} = 1$.The intercepts on the axes are $X_{int} = \sqrt{a}\sqrt{x_1}$ and $Y_{int} = \sqrt{a}\sqrt{y_1}$.The sum of the intercepts is $\sqrt{a}(\sqrt{x_1} + \sqrt{y_1}) = \sqrt{a}(\sqrt{a}) = a$.

The sum of the intercepts on the axes of the tangent to the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ is equal to:

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