Given that M is the point on the ellipse x ^ 2 / 5 + y ^ 2 / 4 = 1, F1 and F2 are the two focal points of the ellipse. If the angle f1mf2 = 30 °, try to find the area of triangle mf1f2

Given that M is the point on the ellipse x ^ 2 / 5 + y ^ 2 / 4 = 1, F1 and F2 are the two focal points of the ellipse. If the angle f1mf2 = 30 °, try to find the area of triangle mf1f2

So, we want to see a 178; (5a = 5A = 5B and 178; (4C; and we want to get 178; (; (4C; and we want to get 178; and we want to get 178; and we want to get 178; and we want to get 178; (1c = 1C = 1mf1 + MF2 = 2A = 2 √ 5mf1 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\lett = M

It is known that the ellipse x ^ 2 / 25 + y ^ 2 / 16 = 1, F1 and F2 are the focal points, and M is the point on the ellipse, if the area of the inscribed circle of △ mf1f2 is 9 π / 4,
Then the number of such points M

If the area of the inscribed circle of △ mf1f2 is 9 π / 4, the radius of the inscribed circle is 3 / 2
The area of the triangle is equal to the half perimeter * the radius of the inscribed circle. The half perimeter of the triangle is (2a + 2C) / 2 = a + C = 8, so the area is 12
The area of the triangle is also equal to the half focal length * height, and the height is 4, so the m point can only be the endpoint of the minor axis, and there are two

The locus of the center of gravity of △ pf1f2 is ()
A. An ellipse with the same eccentricity as B. an ellipse with a different eccentricity from B
C. An ellipse (without the two ends of the major axis) with the same eccentricity as
D. An ellipse (with the two ends of the major axis removed), but with a different eccentricity

If the coordinates of P are (x, y) and the coordinates of gravity center are (x / 3, Y / 3), let the elliptic equation be (x / a) ^ 2 + (Y / b) ^ 2 = 1 eccentricity e = √ (a ^ 2-B ^ 2) / a locus equation of gravity center: (3x / a) ^ 2 + (3Y / b) ^ 2 = 1 eccentricity e = √ (a