The sum of intercepts on coordinate axes made by tangent to the curve $\sqrt{x}+\sqrt{y}=\sqrt{a}$ is

The sum of intercepts on coordinate axes made by tangent to the curve $\sqrt{x}+\sqrt{y}=\sqrt{a}$ is
  1. a
  2. 2 a
  3. $2 \sqrt{\mathrm{a}}$
  4. $\sqrt{2} a$

Solution

$\sqrt{x}+\sqrt{y}=\sqrt{a}$
Differentiating w.r.t. $x$, we get $\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{\mathrm{~d} y}{\mathrm{~d} x}=0$ $\therefore \quad$ Slope of the tangent is $\frac{\mathrm{d} y}{\mathrm{~d} x}=-\sqrt{\frac{y}{x}}$ $\therefore \quad$ Equation of the tangent at $\left(x_1, y_1\right)$ is $\left(y-y_1\right)=-\sqrt{\frac{y_1}{x_1}}\left(x-x_1\right)$ $\begin{aligned} & \sqrt{x_1} y+\sqrt{y_1} x=\sqrt{x_1} y_1+\sqrt{y_1} x_1 \\ \therefore \quad & \sqrt{y_1} x+\sqrt{x_1} y=\sqrt{x_1 y_1}\left(\sqrt{x_1}+\sqrt{y_1}\right) \\ \therefore \quad & \sqrt{y_1} x+\sqrt{x_1} y=\sqrt{x_1 y_1 \mathrm{a}} \\ \therefore \quad & \frac{x}{\sqrt{x_1 \mathrm{a}}}+\frac{y}{\sqrt{y_1 \mathrm{a}}}=1 \\ \therefore \quad & x \text {-intercept }+y \text {-intercept }=\sqrt{\mathrm{a}}\left(\sqrt{x_1}+\sqrt{y_1}\right) \\ & =\sqrt{\mathrm{a}} \times \sqrt{\mathrm{a}}=\mathrm{a}\end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 1)

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