# Family of Plane Curves

Find the differential equation of parabolas with vertex and focus on the x-axis.

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

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