Question

The sum of 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

Answer: (D)

Given, curve,
$\sqrt{x}+\sqrt{y}=3$
On differentiating with respect to $x$, we get
$\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0$
$\frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(4,1\right)}=-\frac{\sqrt{1}}{\sqrt{4}}=\frac{-1}{2}\left[\therefore x_1=4,y_1=1\right]$
Equation of tangent of curve at $\left(4,1\right)$
$y-1=\frac{-1}{2}\left(x-4\right)$
$\Rightarrow x+2y=6$
$\Rightarrow \frac{x}{6}+\frac{y}{3}=1$
Sum of intercept $6+3=9$