Question

The equation of the tangent to curve $\sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=2$ at the point (a, b) is

• A. $\frac{x}{a}-\frac{y}{b}=0$
• B. $\frac{x}{a}+\frac{y}{b}=2$
• C. $\frac{x}{a}-\frac{y}{b}=1$
• D. $\frac{x}{a}+\frac{y}{b}=0$

Answer 1

Answer: (B)

$\frac{x}{a}+\frac{y}{b}=2$

We have,

$\sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=2.......\left(1\right)$

$\frac{\sqrt{x}}{\sqrt{a}}+\frac{\sqrt{y}}{\sqrt{b}}=2$

On differentiation and we get,

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

$\frac{1}{2\sqrt{y}\sqrt{b}}\frac{dy}{dx}=-\frac{1}{2\sqrt{x}\sqrt{a}}$

$\frac{dy}{dx}=-\frac{\sqrt{y}\sqrt{b}}{\sqrt{x}\sqrt{a}}$

At the point $(a,b)$ and we get,

$\frac{dy}{dx}=-\frac{\sqrt{b}\sqrt{b}}{\sqrt{a}\sqrt{a}}$

$\frac{dy}{dx}=-\frac{b}{a}$

Equation of tangent is

$y-y_1=\frac{dy}{dx}(x-x_1)$

$y-b=-\frac{b}{a}(x-a)$

$ay-ab=-bx+ab$

$ay+bx=2ab$

$\frac{ay+bx}{ab}=2$

$\frac{y}{b}+\frac{x}{a}=2$

$\frac{x}{a}+\frac{y}{b}=2$

Hence, this is the answer.