JEE Main 2021 (Online) 3rd August Morning ShiftQuestion 1
The sum of the intercepts on the coordinate axes made by the tangent at any point on the curve, $\sqrt{x}+\sqrt{y}=3$ is
Differentiation and its Application
Solution
Concept. Find the tangent at $(x_0,y_0)$ and its intercepts; they simplify using $\sqrt{x_0}+\sqrt{y_0}=3$. Formulas used. (1) $\dfrac{1}{2\sqrt x}+\dfrac{y'}{2\sqrt y}=0$ $\Rightarrow$ $y'=-\sqrt{\dfrac yx}$Step 1 - Tangent. By (1), $y-y_0=-\sqrt{\frac{y_0}{x_0}}\,(x-x_0)$. Step 2 - $x$-intercept. Put $y=0$: $x=x_0+\sqrt{x_0y_0}=\sqrt{x_0}\left(\sqrt{x_0}+\sqrt{y_0}\right)=3\sqrt{x_0}$. Step 3 - $y$-intercept. In the same way, $3\sqrt{y_0}$. Step 4 - Sum. $3\left(\sqrt{x_0}+\sqrt{y_0}\right)=9$. Answer: (C)