Question

The sum of the intercepts made by a tangent to the curve $\sqrt{x}+\sqrt{y}=4$ at point $(4,4)$ on coordinate axes is -

• A. $4\sqrt{2}$
• B. $6\sqrt{3}$
• C. $8\sqrt{2}$
• D. $\sqrt{256}$

Answer 1

The correct option is D $\sqrt{256}$

$\sqrt{x}+\sqrt{y}=4$

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

$\Rightarrow \left(\frac{dy}{dx}\right)=-\sqrt{\frac{y}{x}}$

Thus slope of tangent at $(4,4)$ is $={\left(\frac{dy}{dx}\right)}_{(4,4)}=-1$

The equation of tangent is $(y-4)=-1(x-4)\Rightarrow x+y=8$

Thus, sum of intercept made by tangent on the axis is $=8+8=16=\sqrt{256}$

Hence, option 'D' is correct.