Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 42

Suppose that the curve has equation of $\sqrt{x} + \sqrt{y} = \sqrt{c}$. Show that the sum of the $x$ and $y$ intercepts of any tangent line to the curve is equal to $c$.

Taking the derivative of the curve implicitly we have.


$
\begin{equation}
\begin{aligned}

\frac{1}{ 2 \sqrt{x}} + \frac{\displaystyle \frac{dy}{dx}}{2 \sqrt{y}} =& 0
\\
\\
\frac{dy}{dx} =& -\frac{\sqrt{y}}{\sqrt{x}}

\end{aligned}
\end{equation}
$


Using Point Slope Form


$
\begin{equation}
\begin{aligned}

y - y_1 =& m (x - x_1)
\\
\\
y - y_1 =& - \frac{\sqrt{y_1}}{\sqrt{x_1}} (x - x_1)

\end{aligned}
\end{equation}
$


Recall that $x$-intercept is the value of $x$ where $y = 0$ consequently, $y$-intercept is the value of $y$ where $x = 0$. So,


$
\begin{equation}
\begin{aligned}

\text{For } x =& 0,
\\
\\
y - y_1 =& - \frac{\sqrt{y_1}}{\sqrt{x_1}} (0 - x_1)
\\
\\
y =& y_1 - \frac{\sqrt{y_1}}{\sqrt{x_1}} (-x_1)
\\
\\
y =& y_1 + \sqrt{y_1} \sqrt{x_1}
\\
\\
\text{For } y =& 0,
\\
\\
0 - y_1 =& - \frac{\sqrt{y_1}}{\sqrt{x_1}} (x - x_1)
\\
\\
x - x_1 =& \frac{y_1 \sqrt{x_1}}{\sqrt{y_1}}
\\
\\
x =& \frac{y_1 \sqrt{x_1}}{\sqrt{y_1}} + x_1
\\
\\
x =& x_1 + \sqrt{x_1} \sqrt{y_1}

\end{aligned}
\end{equation}
$


Thus, the $x$ and $y$-intercepts are $x = x_1 + \sqrt{x_1} \sqrt{y_1}$ and $y = y_1 - \sqrt{x_1} \sqrt{y_1}$

Now, solving for the condition...


$
\begin{equation}
\begin{aligned}

(x_1 + \sqrt{x_1} \sqrt{y_1}) + (y_1 \sqrt{x_1} \sqrt{y_1}) =& c
\\
\\
x_1 + 2 \sqrt{x_1} \sqrt{y_1} + y_1 =& c
\\
\\
\text{but } \sqrt{c} =& \sqrt{x} + \sqrt{y}
\\
\\
c =& (\sqrt{x} + \sqrt{y})^2
\\
\\
c =& x + \sqrt{x} \sqrt{y} + \sqrt{x} \sqrt{y} + y
\\
\\
c =& x + 2 \sqrt{x} \sqrt{y} + y

\end{aligned}
\end{equation}
$


Therefore, it shows that the sum of the $x$ and $y$-intercepts of any tangent line to the curve is equal to $c$.