Question

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

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

Answer 1

Answer: (B)

$a$

Let $\sqrt{x}+\sqrt{y}=\sqrt{a}$

diff. curt.

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

$\frac{dx}{dy}=-\sqrt{\frac{y}{x}}$ (slope of tangent)

To find points,

Let $x^{\circ }=c$

$y^{\circ }={(\sqrt{a}-\sqrt{c})}^2$

So, equation of tangent ,

$\frac{y-{(\sqrt{a}-\sqrt{c})}^2}{x-c}=-\sqrt{\frac{y}{x}}(=\frac{\sqrt{c}-\sqrt{a}}{\sqrt{c}})⟶\left(1\right)$

Intercept of x,

put $y=0$

$x=\sqrt{ac}⟶\left(2\right)$

For intercept of y,

put $x=0$ in equation(1)

We get, $y=a-\sqrt{ac}⟶\left(3\right)$

Sum of both (2)+(3)

$a-\sqrt{a}c+\sqrt{a}c$

$\Rightarrow a$