So the fixed points are $$F_1=(p_1,q_1)$$$$F_2=(p_2, q_2)$$ Mid-point of foci(centre) is $$\left(\cfrac{p_1+p_2}{2},\cfrac{q_1+q_2}{2}\right)=(c_x,c_y)$$ and the the point $P=(h,k)$
The equation is given by:
$$|PF_1-PF_2|=d$$
Further solving results in :
Case:1
$$-2(p_1-p_2)h-2(q_1-q_2)k+q_1^2-q_2^2+p_1^2-p_2^2=d^2+2d\sqrt{(p_2-h)^2+(q_2-k)^2}.$$
Case:2
$$-2(p_2-p_1)h-2(q_2-q_1)k+q_2^2-q_1^2+p_2^2-p_1^2=d^2+2d\sqrt{(p_2-h)^2+(q_2-k)^2}.$$
Combining the results
$$(p_1-h)^2+(q_1-k)^2+(p_2-h)^2+(q_2-k)^2-2\left(\sqrt{(p_1-h)^2+(q_1-k)^2}.\sqrt{(p_2-h)^2+(q_2-k)^2}\right)=d^2$$
Is there any further simplification that can be done?
In hindsight this has to be the equation of hyperbola. So I proceed to change it to such.
Let coordinates of the vertices of hyperbola be $(v_x,v_y)$.
Then I think$$\frac{v_y-c_y}{v_x-c_x}=\frac{v_y-q_1}{v_x-p_1}=\frac{q_2-q_1}{p_2-p_1}$$
However I don't know How to proceed further?
This proof assumes the center of hyperbola to be at (0,0) and x-axis to be transverse axis. I was trying for a more general equation.