The sum of the intercepts on the coordinate a | vedclass

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Question

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

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$a$

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(A) Given the curve equation: $\sqrt{x} + \sqrt{y} = \sqrt{a}$.Differentiating with respect to $x$,we get: $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0$.Thus,$\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}$.The equation of the tangent at point $(x_0, y_0)$ is $Y - y_0 = -\frac{\sqrt{y_0}}{\sqrt{x_0}}(X - x_0)$.Simplifying,$X\sqrt{y_0} + Y\sqrt{x_0} = x_0\sqrt{y_0} + y_0\sqrt{x_0} = \sqrt{x_0y_0}(\sqrt{x_0} + \sqrt{y_0}) = \sqrt{x_0y_0}\sqrt{a}$.Dividing by $\sqrt{a}\sqrt{x_0y_0}$,we get the intercept form: $\frac{X}{\sqrt{a}\sqrt{x_0}} + \frac{Y}{\sqrt{a}\sqrt{y_0}} = 1$.The intercepts on the axes are $X_{int} = \sqrt{a}\sqrt{x_0}$ and $Y_{int} = \sqrt{a}\sqrt{y_0}$.The sum of the intercepts is $\sqrt{a}(\sqrt{x_0} + \sqrt{y_0}) = \sqrt{a}(\sqrt{a}) = a$.

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

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