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) $
$ ⇒ x +2 y = 6 $
$ \Rightarrow \frac{x}{6} +\frac{y}{3} = 1 $
Sum of intercept $6+3 = 9 $