Question

The sum of the intercepts on the axes of the tangent to the curve $\sqrt{x}+\sqrt{y}=3$ at $(4,1)$ is

• A. $3$
• B. $5$
• C. $7$
• D. $9$

Answer 1

The correct option is D $9$

$\sqrt{x}+\sqrt{y}=3$
Differentiating w.r.t. $x$
$\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\frac{dy}{dx}=0$
Putting $(4,1)$
$\Rightarrow \frac{dy}{dx}{|}_{(4,1)}=m=-\frac{1}{2}$
Equation of tangent:
$\Rightarrow y-1=\frac{-1}{2}(x-4)\Rightarrow \frac{y}{3}+\frac{x}{6}=1$

Hence the sum of intercepts $=9$