Tangent Line Intercept

Question

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.

Answer 1

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}

Answer 2

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.

Answer 3

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 $x_0+\sqrt{x_0y_0}+y_0+\sqrt{x_0y_0}=(\sqrt{x_0}+\sqrt{y_0})^2=c.$