Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 14 / c / iii / Solution

From $p^2=C/(x+C)$, we have $1-p^2=x/(x+C)$. Substitution into the original relation gives $y=2(x+C)p$, soThese are real parabola branches, with either sign of $C$ allowed when the branch is real. For $C\ne 0$, differentiating gives $p=2C/y$ and verifies both the relation and the slope. Use graph intervals with $y\ne 0$ and $x\ne 0$, so the slope is finite and $p\ne \pm 1$. 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.