Let there be two parabolas with the same axis, focus of each being exterior to the other and the latus rectum being 4a and 4b. The locus of the middle points of the intercepts between the parabolas made on the lines parallel to the common axis is a

(a, b) With choice of axes as shown in the figure, the equations of the parabolas can be taken as $y^2=4a\left(x-k\right)$ and $y^2=-4b\left(x+k\right)$.

A line parallel to the common axis is y = h.

Then $A=\left(\frac{h^2}{4a}+k, h\right)$ and $B=\left(-k-\frac{h^2}{4b}, h\right)$.

$\therefore$ if $P=\left(\alpha ,\beta \right)$ be midpoint of AB then

$\alpha =\frac{1}{2}\left(\frac{h^2}{4a}+k-k-\frac{h^2}{4b}\right),\beta =h\therefore 2\alpha =\frac{h^2}{4}\left(\frac{1}{a}-\frac{1}{b}\right)$

$\therefore$ the locus of P is $2x=\frac{y^2}{4}\left(\frac{b-a}{ab}\right)$