Since the vertex and focus are on the $x-axis$, then their coordinates are in the form $(a,0)$ and $(b,0)$ respectively. The general equation of a parabola with axis of symmetry at the $x-axis$ is $(x-a)^2 = 4ky$. There will be two arbitrary constants since there is no fixed vertex/focus and length of latus rectum.
$$\begin{eqnarray}
(x-a)^2 &=& 4ky\\
(x-a)^2 y^(-1) &=& 4k\\
2(x-a)y^(-1) - (x-a)^2 y^(-2)y'&=& 0\\
2y - (x-a)y' &=& 0\\
x - \dfrac{2y}{y'} &=& a\\
1 - \dfrac{2y' y' - 2yy''}{(y')^2} &=& 0\\
(y')^2 - (2(y')^2 - 2yy'') &=& 0\\
-(y')^2 + 2yy'' &=& 0\\
(y')^2 - 2yy'' &=& 0 \rightarrow Answer
\end{eqnarray}$$
Forum posts (unless otherwise specified) licensed under a Creative Commons Licence. All trademarks and copyrights on this page are owned by their respective owners. Forum posts are owned by the individual posters.
Since the vertex and focus
Since the vertex and focus are on the $x-axis$, then their coordinates are in the form $(a,0)$ and $(b,0)$ respectively. The general equation of a parabola with axis of symmetry at the $x-axis$ is $(x-a)^2 = 4ky$. There will be two arbitrary constants since there is no fixed vertex/focus and length of latus rectum.
$$\begin{eqnarray}
(x-a)^2 &=& 4ky\\
(x-a)^2 y^(-1) &=& 4k\\
2(x-a)y^(-1) - (x-a)^2 y^(-2)y'&=& 0\\
2y - (x-a)y' &=& 0\\
x - \dfrac{2y}{y'} &=& a\\
1 - \dfrac{2y' y' - 2yy''}{(y')^2} &=& 0\\
(y')^2 - (2(y')^2 - 2yy'') &=& 0\\
-(y')^2 + 2yy'' &=& 0\\
(y')^2 - 2yy'' &=& 0 \rightarrow Answer
\end{eqnarray}$$