Question

The sum of the intercepts of a tangent to $\sqrt{x}+\sqrt{y}=\sqrt{a},a>0$ upon the coordinate axes is

• A. $2a$
• B. $a$
• C. $a/2$
• D. $\sqrt{a}$

Answer 1

Answer: (B)

$a$

$\sqrt{x}+\sqrt{y}=\sqrt{a},a>0$

Differentiating w.r.t $x$, we get

$\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}$

Equation of tangent at any point $(x,y)$ of the curve is

$Y-y=-\frac{\sqrt{y}}{\sqrt{x}}(X-x)$

So intercepts of x-axis and y-axis are $x+\sqrt{xy}$ and $y+\sqrt{xy}$.

Therefore, the sum of intercepts

$=x+y+2\sqrt{xy}={(\sqrt{x}+\sqrt{y})}^2=a$