A Tangent Line Problem

Let \(\sqrt{x}+\sqrt{y}=\sqrt{c}\) describe a curve for some \(c>0\). Prove that for any tangent line, the sum of \(x\) and \(y\) intercepts is \(c\). This is problem 74 in section 2.5 of Calculus of a Single Variable 8th Edition by Larson, Hostetler, and Edwards.

First, find the derivative (implicitly) and write it in terms of \(x\) only.

\[\begin{align}&\rx+\ry=\rc\\&{1\over2\rx}+{y'\over2\ry}=0\\& y'={-\ry\over\rx}={\rx-\rc\over\rx}\end{align}\]

Now let \(x=x'\) and we will solve for \(y\), which gives us the point \((x',y')\) on the curve.

\[\ryp=\rc-\rxp\Rightarrow y'=c+x'-2\rcxp\]

Then form the equation for the tangent line using point slope form.

\[\begin{align}& y-(c+x'-2\rcxp)={\rxp-\rc\over\rxp}(x-x') \end{align}\]

After simplifying, the equation for the line is this.

\[y={\rxp-\rc\over\rxp}x+c-\rcxp\]

The \(y\) intercept is \(c-\rcxp\). The \(x\) intercept is.

\[\begin{align}&{-(c-\rcxp)\over(\rxp-\rc)/\rxp} ={-(c-\rcxp)\rxp\over\rxp-\rc}={x'\rc-c\rxp\over\rxp-\rc} \cdot{\rxp+\rc\over\rxp+\rc}\\& ={x'\rcxp-cx'+cx'-c\rcxp\over x'-c} ={\rcxp(x'-c)\over x'-c}=\rcxp \end{align}\]

So when both intercepts are added, we get \(c\).