The question is like this: there is a curve $\sqrt{x}+\sqrt{y}=\sqrt{c}$, where $c$ is a real number and $c > 0$. If $L$ is a tangent line, any tangent line, with only 1 x-intercept and 1 y-intercept, prove that the sum of the x, y-intercepts of $L$ is $c$.
First of all, I used implicit differenciation on the equation. $$\sqrt{x}+\sqrt{y}=\sqrt{c}$$
Then I also put a graph here
(The curve when c=4)
But I still do not have idea. Thank you for looking at this question.
Best Answer
From implicit differentiation you get $y'=-\sqrt{\frac{y}{x}}$. Hence the tangent line at any point $(x_0,y_0)$ is given by $y=-\sqrt{\frac{y_0}{x_0}}(x-x_0)+y_0$. Now find the intercepts and check that there sum is $x_0+\sqrt{x_0y_0}+y_0+\sqrt{x_0y_0}=(\sqrt{x_0}+\sqrt{y_0})^2=c.$