Conic In the plane rectangular coordinate system, there are two fixed points F1 (0, radical 3) F2 (0, - radical 3). If the moving point m satisfies MF1 + MF2 = 4 Let the line L: y = KX + T intersect the curve with two points a and B, and let the line L1: y = K1X intersect at point D. if K × K1 = - 4, it is proved that D is the midpoint of ab

Conic In the plane rectangular coordinate system, there are two fixed points F1 (0, radical 3) F2 (0, - radical 3). If the moving point m satisfies MF1 + MF2 = 4 Let the line L: y = KX + T intersect the curve with two points a and B, and let the line L1: y = K1X intersect at point D. if K × K1 = - 4, it is proved that D is the midpoint of ab

The trajectory of M is ellipse 2A = 4A = 2, a & sup2; = 4C = √ 3, so B & sup2; = A & sup2; - C & sup2; = 1, the focus is on the y-axis, so the equation is X & sup2; + Y & sup2 / 4 = 1. Let a (x1, Y1) B (X2, Y2) substitute the coordinates of a and B into the elliptic equation, and then subtract them, and divide them by x1-x2 at the same time

Construction of conic
We know that F1 and F2 are
Square of elliptic equation x / 4 + square of Y / 3 = 1
The left and right focus, AB, is over the right focus, F2
The maximum area of the inscribed circle of △ Abf1 is

I don't know if it's a big question or a blank question,
Because the circumference of the triangle Abf1 is L = 4A = 8,
The radius of the inscribed circle r = 2S △ Abf1 / L,
So when the area of △ ABF 1 is the largest, the radius of the inscribed circle is the largest
If and only if, ab ⊥ X axis, s △ Abf1 is the largest, and the area of inscribed circle is the largest
In this case, s △ Abf1 = (1 / 2) (2) * 3 = 3
So r (max) = 2 * 3 / 8 = 3 / 4
The maximum area of inscribed circle is π (3 / 4) ^ 2 = 9 π / 16