PF 1^2+PF2^2=F 1F2^2，
Let the elliptic focal coordinates F 1(-c, 0), F2(c, 0),
|F 1F2|=2c，
c=√(a^2-b^2)，
pf 1^2+pf2^2=4c^2=4(a^2-b^2)，
|PF 1|+|PF2|=2a，
(|PF 1|+|PF2|)^2=4a^2，
pf 1^2+pf2^2+2|pf 1|*|pf2|=4a^2，
4(a^2-b^2)+2|pf 1|*|pf2|=4a^2，
|PF 1|*|PF2|=2b^2，
|pf 1|(2a-|pf 1|)=2b^2，
|pf 1|^2-2a|pf 1|+2b^2=0，
To make the equation meaningful, △ = 4a2-8b2 >; =0,
a^2>； =2b^2，
∴a≥√2b，
Make a circle with the origin as the center and c as the radius, if a >；; 2b, there are four intersections, there are four such triangles, one vertex in each quadrant.
When b=c, the ellipse is tangent to the circle, and at the two ends of the short axis, * * * has two intersections, so there are two such triangles.