Answer:
PA.PB=600
Step-by-step explanation:
Given Data
Major axis=50
Minor axis=40
conjugate axis of length=20
P be a point on both the hyperbola and ellipse
PA.PB=?
Solution
Taken definition of ellipse
PA+PB=50
Minor axis of ellipse is 40 then a=25 and b=20
Which makes focal distance c=15
For hyperbola
[tex]a^{2}+b^{2}=c^{2}[/tex]
2b is conjugate axis
2a is the transverse axis
So for this hyperbola 2b=20
b=20/2
b=10
and c=15
then
[tex]a=\sqrt{c^{2}-b^{2} }\\ a=\sqrt{(15)^{2} -(10)^{2} } \\a=\sqrt{25}\\ a=5[/tex]
that makes
2a=10 Distance between vertices
If P is point
So for this hyperbola
|PA-PB|=10
[tex](PA-PB)^{2} =100[/tex]
We know for ellipse
PA+PB=50
[tex](PA+PB)^{2}=2500[/tex]
Since
[tex](PA+PB)^{2}-(PA-PB)^{2}\\ 4PA.PB=2500-100\\PA.PB=\frac{2400}{4}\\ PA.PB=600[/tex]