Question

The sum of the intercepts made on the axes of coordinates by any tangent to the curve $\sqrt{x}+\sqrt{y}=2$ is equal to

• A. $4$
• B. $2$
• C. $8$
• D. None of these

Answer 1

The correct option is A $4$

Curve equation $\sqrt{x}+\sqrt{y}=2\Rightarrow x>0,y>0$
$\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0$
$\frac{dy}{dx}=-\sqrt{\frac{y}{x}}$
and since in question its asking in general not for a particular point on curve. Hence it we find it for any point on curve then it will be true for any point.
So as use can see (1, 1) lies on curve
So, $\frac{dy}{dx}{\int }_{1,1}=-1$
So tangent equation at (1, 1) is $y=(-1)x+c$
and it will pass through (1, 1)
So, $c=2$
tangent equation : $y=-x+2$
$x+y=2$
$\frac{x}{2}+\frac{y}{2}=1$
So x - intercept $=2$
y- intercept $=2$
So sum $=4$