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.
$$\begin{eqnarray}
(x-a)^2 &=& 4ky\\
(x-a)^2 y^(-1) &=& 4k\\
2(x-a)y^(-1) - (x-a)^2 y^(-2)y'&=& 0\\
2y - (x-a)y' &=& 0\\
x - \dfrac{2y}{y'} &=& a\\
1 - \dfrac{2y' y' - 2yy''}{(y')^2} &=& 0\\
(y')^2 - (2(y')^2 - 2yy'') &=& 0\\
-(y')^2 + 2yy'' &=& 0\\
(y')^2 - 2yy'' &=& 0 \rightarrow Answer
\end{eqnarray}$$

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}$$

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