# Tangent Line Intercept

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.

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 $$x_0+\sqrt{x_0y_0}+y_0+\sqrt{x_0y_0}=(\sqrt{x_0}+\sqrt{y_0})^2=c.$$
Result: Let $$\phi$$ be the curve in the $$xy$$-plane defined by $$\sqrt{x}+\sqrt{y}=\sqrt{c}$$, where $$c\in\mathbb{R}^+$$, and let $$L$$ be a line tangent to $$\phi$$ with exactly one $$x$$-intercept $$x_i$$ and exactly one $$y$$-intercept $$y_i$$. Then $$x_0+y_0=c$$.
Proof: We implicitly differentiate $$\phi$$ to find the slope of $$L$$: \begin{align} \frac{d}{dx}\left[\sqrt{x}+\sqrt{y}\right] &= \frac{d}{dx}\left[\sqrt{c}\right] \\ \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\cdot\frac{dy}{dx} &= 0 \\ \frac{1}{2\sqrt{y}}\cdot\frac{dy}{dx} &= -\frac{1}{2\sqrt{x}} \\ \frac{dy}{dx} &= -\sqrt{\frac{y}{x}}. \end{align} Given any point $$(x_0,y_0)$$ on $$\phi$$, the slope of $$L$$ through $$(x_0,y_0)$$ is $$\frac{dy}{dx}\bigg|_{(x_0,y_0)}=-\sqrt{\frac{y_0}{x_0}}$$ and so the rectangular equation for $$L$$ is \begin{align} y-y_0 &= -\sqrt{\frac{y_0}{x_0}}\left(x-x_0\right) \\ \frac{y}{\sqrt{y_0}}-\frac{y_0}{\sqrt{y_0}} &= -\frac{x}{\sqrt{x_0}}+\frac{x_0}{\sqrt{x_0}} \\ \frac{y}{\sqrt{y_0}}+\frac{x}{\sqrt{x_0}} &= \sqrt{x_0}+\sqrt{y_0} \\ \frac{x}{\sqrt{x_0}}+\frac{y}{\sqrt{y_0}} &= \sqrt{c}.\tag{Since (x_0,y_0) lies on \phi} \end{align} We now set $$y=0$$ and $$x=0$$ to find the two intercepts of $$L$$, $$x_i$$ and $$y_i$$ respectively. Observe that \begin{align} \frac{x_i}{\sqrt{x_0}}+0 &= \sqrt{c} \\ x_i &= \sqrt{c\,x_0}, \end{align} and \begin{align} 0+\frac{y_i}{\sqrt{x_0}} &= \sqrt{c} \\ y_i &= \sqrt{c\,y_0}. \end{align} Finally, we sum $$x_i$$ and $$y_i$$, which yields \begin{align} x_i+y_i &= \sqrt{c\,x_0}+\sqrt{c\,y_0} \\ &= \sqrt{c}\left(\sqrt{x_0}+\sqrt{y_0}\right) \\ &= \sqrt{c}\cdot\sqrt{c} \\ &= c.\tag*{\blacksquare} \end{align}
Suppose tangent line intersect the curve at $$(x_{0},y_{0})$$. Then the line equation will be
$$\frac{x}{\sqrt{x_{0}}}+\frac{y}{\sqrt{y_{0}}}=\sqrt{c}$$
Were you able to get this line equation? From here you substitute $$x=0$$ to find $$y$$ intercept and vise versa.