Solution to the second problem: a=3√5, b=2√5, c=5, F 1, the coordinates of F2 are (5,0), (-5,0), and the coordinates of point p are (x, y).

PF 1+PF2=2a=6√5

pf 1^2+pf2^2=2a=(2c)^2= 100

PF 1=4√5 or PF 1=2√5, i.e.

PF 1=a-ex=3√5-√5x/3=4√5 or PF 1=a-ex=3√5-√5x/3=2√5.

X=3 or x=-3.

So the coordinates of point P are (3,4) (3,4) (-3,4) (-3,4) (-3,4).