If P is a point on the ellipse, the vector Pf1 × vector PF2 =? Needs a general conclusion, and F1F2 is the focus

If P is a point on the ellipse, the vector Pf1 × vector PF2 =? Needs a general conclusion, and F1F2 is the focus

Vector Pf1 × vector PF2 = 2B ^ 2 - | Pf1 | * | PF2|

Let F1F2 be the focus of the ellipse x2 / 9 + Y2 / 4 = 1, p be on the ellipse, and | vector Pf1 + vector PF2 | = 2 follow sign 5, then the angle between vector PF2 and Pf1
It's better to be careful in the process,

A = 3 B = 2 C = radical 5
|F1F2 | = 2C = 2 radical 5
|F1F2|=|PF1-PF2|=|PF1+PF2|
The diagonals of the quadrilateral are equal and the corresponding sides are parallel, so Pf1 ⊥ PF2
90 ° angle

Given that P is a point on the ellipse X & # 178 / 4 + Y & # 178; = 1, F1F2 is the ellipse and obtains two focal points, and ∠ f1pf2 = 60 °, find △ F12
Half an hour. Urgent

Let | Pf1 | = m, | PF2 | = n
∵ P on an ellipse
∴m+n=2a=4 ①
∵∠F1PF2=60º,|F1F2|=2c=2√3
According to the cosine theorem:
m²+n²-2mncos60º=4c²
That is M & # 178; + n & # 178; - Mn = 12
①²-②:
3mn=4
∴mn=4/3
∴SΔF1PF2=1/2mnsin60º=1/2*4/3*√3/2=√3/3