Question 1074904
If P is a point in an ellipse with foci A and B, and major axis 50,
by the definition of ellipse,
PA+PB=50.
If the minor axis of that ellipse is 40,
them the semimajor and senior acrs are respectively
a=25 and b=20.
That makes the focal distance
c=15.
 
For a hyperbola, the focal distance,
the distance between the vertices (the transverse axis),
and the conjugate axis are related by
{{{a^2+b^2=c^2}}} , where c is the focal distance,
2b is the conjugate axis, and
2a is the transverse axis.
So for this hyperbola, 2b=20, so b=10,
and c=15, so
{{{a=sqrt(15^2-10^2)=sqrt(225-100)=sqrt(25)=5}}} .
That makes 2a=10.
That is the distance between the vertices.
 
If P is a point in a hyperbola with foci A and B,
{{{abs(PA-PB)}}}= distance between the vertices.
So, for this hyperbola
{{{abs(PA-PB)=10}}} ---> {{{(PA-PB)^2=100}}}
 
We knew for the ellipse that
PA+PB=50 ---> {{{(PA+PB)^2=2500}}} .
 
Since {{{4PA*PB=(PA+PB)^2-(PA-PB)^2}}} ,
{{{4PA*PB=2500-100=2400}}} 
{{{PA*PB=2400/4}}} and {{{highlight(PA*PB=600)}}}