The equation of the tangent to the curve sqrt | vedclass

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(B) 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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$\frac{x}{\sqrt{x_1}} + \frac{y}{\sqrt{y_1}} = \sqrt{a}$

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(B) 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,$\frac{dy}{dx} = -\sqrt{\frac{y}{x}}$.At the point $(x_1, y_1)$,the slope of the tangent is $m = -\sqrt{\frac{y_1}{x_1}}$.The equation of the tangent is $y - y_1 = -\sqrt{\frac{y_1}{x_1}}(x - x_1)$.Multiplying by $\sqrt{x_1}$,we get $y\sqrt{x_1} - y_1\sqrt{x_1} = -x\sqrt{y_1} + x_1\sqrt{y_1}$.Rearranging terms,$x\sqrt{y_1} + y\sqrt{x_1} = x_1\sqrt{y_1} + y_1\sqrt{x_1}$.Since $(x_1, y_1)$ lies on the curve,$\sqrt{x_1} + \sqrt{y_1} = \sqrt{a}$.Dividing the equation $x\sqrt{y_1} + y\sqrt{x_1} = \sqrt{x_1}\sqrt{y_1}(\sqrt{x_1} + \sqrt{y_1})$ by $\sqrt{x_1}\sqrt{y_1}$,we get $\frac{x}{\sqrt{x_1}} + \frac{y}{\sqrt{y_1}} = \sqrt{x_1} + \sqrt{y_1} = \sqrt{a}$.

The equation of the tangent to the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ at the point $(x_1, y_1)$ is:

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