A variable parabola of latus rectum $l $, touches a fixed equal parabola , the axes of the two curves being parallel . Then locus of the vertex of the moving curve is a parabola , then what is the latus rectum of this curve .
I took the fixed curve as $y^2 =4bx$
And the variable curve is $(y-k)^2 = 4a(x-h)$
Where $(h,k)$ is the vertex . But now how to proceed ?
I like projective geometry, so I'll be explaining using terms from there. Your first parabola can be written as
$$(x,y,1)\cdot\begin{pmatrix}0&0&2b\\0&-1&0\\2b&0&0\end{pmatrix} \cdot\begin{pmatrix}x\\y\\1\end{pmatrix}=0$$
Its axis is the $x$ axis, evidenced by the fact that $(1,0,0)$ is the only point at infinity (i.e. last coordinate zero) which lies on that conic as well. Now I'd like to describe the tangent for a given direction. A direction can be described by a point at infinity. I'll use the point $(t,1,0)$ with parameter $t$. For $t\in\mathbb R$ this can describe any direction except horizontal, which is just fine since that parabola can't have a horizontal tangent in any case.
So we need the tangent through a given point. One way to compute that is by computing the polar line of that point. It will intersect the conic in the two contact points. Usually that leads to two tangents, but in our case, with the point chosen at infinity, one of the tangents will be the line at infinity, so we can pick the remaining (finite) contact point.
To compute a polar line, multiply the point with the matrix:
$$\begin{pmatrix}0&0&2b\\0&-1&0\\2b&0&0\end{pmatrix} \cdot\begin{pmatrix}t\\1\\0\end{pmatrix}= \begin{pmatrix}0\\-1\\2bt\end{pmatrix}$$
This is the line $y=2bt$. It intersects the conic at $x=\frac{y^2}{4b}=bt^2$. So $(bt^2,2bt)$ is the point with the given tangent direction. Now do the same for the second curve with $a$ instead of $b$. Setting the point of contact equal, you get
\begin{align*} bt^2 &= at^2+h \\ 2bt &= 2at+k \end{align*}
You can eliminate the variable $t$ from this, e.g. using the resultant computed using the Sylvester matrix
$$0=\begin{vmatrix} a-b & 0 & h \\ 2(a-b) & k & 0 \\ 0 & 2(a-b) & k \end{vmatrix}=(a-b)(k^2 + 4ah - 4bh)$$
So the equation of the locus of the vertex is
$$k^2 = 4(b-a)h$$