Question

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

• A. a
• B. 2a
• C. $2\sqrt{a}$
• D. None of these

Answer 1

The correct option is A a

$\sqrt{x}+\sqrt{y}=a;\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0,\therefore \frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}$
Hence tangent at (x, y) is $Y-y=-\frac{\sqrt{y}}{\sqrt{x}}(X-x)$
or $X\sqrt{y}+Y\sqrt{x}=\sqrt{xy}(\sqrt{x}+\sqrt{y})=\sqrt{axy}$
or $\frac{X}{\sqrt{a}\sqrt{x}}+\frac{Y}{\sqrt{a}\sqrt{y}}=1$
Clearly its intercepts on the axes are $\sqrt{a}\sqrt{x}$ and $\sqrt{a}\sqrt{y}$
Sum of the intercepts = $\sqrt{a}(\sqrt{x}+\sqrt{y})=\sqrt{a}.\sqrt{a}=a$