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The question is actually to find the order of the differential equation of this family, but I cannot formulate the equation.

I searched a little and found the equation to be something like $(y-k)^2=4ax$. But if we assume the directrix to be the line x = -a and take a specific case when k=0, then there will be a parabola having vertex at origin ($y^2=4ax$) and another parabola having vertex at (-2a,0) opening on the left side having the same directrix. But that does not come under the equation representing this family.

Am I thinking something wrong?

This is what I mean:

This is what I mean.

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    The second order differential equation and its proof can be found in (https://www.quora.com/What-is-the-equation-for-family-of-parabolas?share=1). – 2017-02-02
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    In their explanation, take at once $h=p$ and express the definition of a parabola as the set of points $M(x,y)$ such that the distance $MH=x$ from the $y$ axis (the directrix) is equal to the distance MF to the focus $F(2p,0)$; in fact we write the equality $MH^2=MF^2$ under the form $x^2=(2p-x)^2+y^2$, giving $y^2=4p(x-p),$ as desired. – 2017-02-02
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    @JeanMarie When the vertex of the parabola is (p,k), where k is any integer, the equation $MH^2=MF^2$ will still hold. Here $MH^2$ will be $x^2$ as before but will $MF^2$ not become $(2p-x)^2+(y-k)^2$? Here too the directrix is the y axis. – 2017-02-02
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    You have to take the vertex always on the $x$-axis. – 2017-02-02
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    @JeanMarie Why? Will a parabola with vertex (p,k) will not have the same directrix as the parabola with vertex (p,0), both having y axis are their directrix. – 2017-02-02
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    @JeanMarie OP has a point here. – 2017-02-02
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    @Narasimhan Thanks. I am going to answer him. – 2017-02-02
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    You are right. You can take the vertex everywhere. I was wrong. – 2017-02-02

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For convenience I assume x-axis as directrix. Variable focus is at $(h,k)$ which should be eliminated by differentiation.

$$ (x-h)^2 + (y-k)^2 =y^2 \tag1 $$ $$ x^2 -2 x h -2 y k + h^2+k^2 =0 \tag2$$ Differentiating with respect to x $$ y^{\prime } k = x-h \tag3$$ Differentiating with respect to x again $$ y^{\prime \prime} =1/k\tag4 $$ Solving for $(h,k)$ for elimination

$$(h,k) = ( x- \frac{ y^{\prime }}{y^{\prime \prime }},\,\frac{ {1 }}{y^{\prime \prime }}) \tag5$$

Plug this into (1) and simplify to get the required second order differential equation of arbitrary focus location. The ODE with x-axis as directrix and axis of symmetry not necessarily coinciding with y-axis is given by:

$$ 2 y y^{\prime\prime} = 1+ y^ {{ \prime}^2} \tag6$$

So now recognizing this curvature ratio property ( may be mentioned in old (Chelsea Publishers?) Conics text-books).. The red curve is the parabola evolute.

The radius of curvature of a parabola is double the normal segment length intercepted by the directrix.

Parabola Directrix Prop

$$ \dfrac{ y y^{\prime\prime}} { 1+ y^ {{ \prime}^2}} = \frac{R_d}{R_c}=\frac12\tag7$$

The offset axis case has been numerically computed and verified by CAS It is shown ( BC to arbitrary start point and slope) at right.

So this property can be used beneficially to define parabolic arc differential equation.

Change of axes gives the ODE of parabola whose axis is parallel to x-axis and displaced arbitrarily parallel with respect to $x-$ axis. The focus in general is not on x-axis.

$$ 2 x y^{\prime\prime} = -y^{\prime}(1+y^{{ \prime}^2}) \tag8$$

This is not preferred ODE in the unsymmetric case as there are two distinct values of $y$ for same $x$, it is a mess.

EDIT1:

In hindsight perhaps the easiest way is to get ODEs of all parabolas on same x-axis directrix is Formula17_MW_WA. Curvature ratio

$$ \frac{PD}{PC}= = \frac{R_d}{R_c}=\dfrac{(a t^2+a) \sec\phi }{2a \sec^{3}\phi} =\frac12=\dfrac{ y y^{\prime\prime}} { 1+ y^ {{ \prime}^2}}. \tag9$$