= Solution
From $p^2=C/(x+C)$, we have $1-p^2=x/(x+C)$. Substitution into the original relation gives $y=2(x+C)p$, so
$$
\boxed{y^2=4C(x+C)}.
$$
These are real <parabola> branches, with either sign of $C$ allowed when the branch is real. For $C\ne0$, differentiating gives $p=2C/y$ and verifies both the relation and the slope. Use graph intervals with $y\ne0$ and $x\ne0$, so the slope is finite and $p\ne\pm1$. The separate constant-slope case $p=0$ gives $y=0$, also represented by the degenerate choice $C=0$. Squaring has not imposed a sign on $y$, so both real graph branches must be included.
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