It is known that the equation of ellipse C is x ^ 2 + 2 (y ^ 2) = 1 Let the left and right focal points of ellipse C be F1 and F2 respectively, and the line L passing through point F1 intersects the ellipse at points m and N. take F2m and f2n as the adjacent sides to make the parallelogram mf2np, and find the maximum length of the diagonal F2P of the parallelogram

It is known that the equation of ellipse C is x ^ 2 + 2 (y ^ 2) = 1 Let the left and right focal points of ellipse C be F1 and F2 respectively, and the line L passing through point F1 intersects the ellipse at points m and N. take F2m and f2n as the adjacent sides to make the parallelogram mf2np, and find the maximum length of the diagonal F2P of the parallelogram

Analysis: it is more convenient to solve this kind of problem with vector. Let the equation of line l be x = MY-1, and X & # 178 / 2 + Y & # 178; = 1, and the simultaneous equation be solved to (M & # 178; + 2) y & # 178; - 2my-1 = 0 △ 0, and let m (x1, Y1) n (X2, Y2), vector F2m = (x1-1, Y1) vector f2n = (x2-1, Y1) vector F2P = direction

The center of ellipse C is at the coordinate origin, the focus is on the x-axis, the coordinate of right focus f is (2,0), and the distance from point F to an end point of the minor axis is √ 6 Qi
Solving the equation of ellipse C

x²/6+y²/2=1

It is known that the two focal points F1F2 of the ellipse x2 / 45 + Y2 / 20 = 1 and the point P (x, y) y > 0 are on the ellipse, so that △ pf1f2 is a right triangle

F1 (- 5,0). F1 (5,0). Set point P as (x, y). Since pf1f2 is a right triangle, Pf1 is perpendicular to PF2. Then we use vector idea, that is, multiplication and addition to get an equation about x.y. then we can solve the value together with the original elliptic equation

Given that point P is the point in the first quadrant of the ellipse X29 + Y25 = 1, F1 and F2 are the two focal points of the ellipse. If the radius of the inscribed circle of △ pf1f2 is 12, then the coordinate of point P is ()
A. （355，2）B. （3114，54）C. （3598，58）D. （2，54）

Let P (x, y) (x, y ＞ 0) ∵ △ pf1f2's area s = 12R (| Pf1 | + | PF2 | + | F1F2 |) = 12Y | F1F2 |. Let P (x, y) (x, y ＞ 0) ∵ △ pf1f2's area s = 12R (| Pf1 | + | PF2 | + | F1F2 |) = 12Y | F1F2 |. Let P (2 × 3 + 2 × 2) = y × 2 × 2, then y = 54