Question 1077688
The ellipse has
{{{a=50/2=25}}} (the semi-major axis), and
{{{b=40/2=20}}} (the semi-minor axis).
The focal distance, {{{c}}} , can be found from
{{{b^2+c^2=a^2}}} .
Substituting known values, {{{20^2+c^2=25^2}}}-->{{{c^2=625-400=225}}}--> {{{c=15}}} . 
So, it looks like this {{{drawing(450,375,-30,30,-25,25,
grid(0),arc(0,0,50,40,0,360),
circle(-15,0,0.5),circle(15,0,0.5),
locate(-15.5,2.5,A),locate(14.5,2.5,B)
)}}} (maybe shifted and/or rotated).
According to the definition of ellipse,
if P is a point of an ellipse with foci A and B,
{{{PA+PB=2a=50}}} .
 
The hyperbola has
{{{c=15}}} , just like the ellipse,
and has {{{b=20/2=10}}} .
Knowing that in a hyperbola
{{{a^2+b^2=c^2}}} , we can find {{{a}}} .
{{{a^2+15^2=10^2}}} 
{{{a^2+225=100}}}
{{{a^2=225-100=125}}}
{{{a=sqrt(125)}}}
In a hyperbola with foci A and B, for any point P, by definition
{{{abs(PA-PB)=2a=2sqrt(125)}}} .
 
Squaring both sides in the equations found involving the distances PA and PB,
{{{(PA+PB)^2=2500}}}-->{{{PA^2+PB^2+2*PA*PB=2500}}} and
{{{(PA-PB)^2=500}}}--->{{{PA^2+PB^2-2*PA*PB=500}}} .
Subtracting {{{PA^2+PB^2-2*PA*PB=500}}} from {{{PA^2+PB^2+2*PA*PB=2500}}} , we get
{{{PA^2+PB^2+2*PA*PB-PA^2-PB^2+2*PA*PB=2500-500}}} --> {{{4*PA*PB=2000}}}  --> {{{PA*PB=2000/4=highlight(500)}}}