It is known that the focus F of the parabola C: y2 = 2px (P & gt; 0) coincides with the right focus of the ellipse x24 + Y23 = 1, the line L crosses the parabola at two points a and B through point F, and the projections of points a and B on the directrix of the parabola C are points D and e respectively. (I) the process of finding the parabola C; (II) if the line L crosses the y-axis at point m, and Ma = MAF, MB = NBF, is the fixed value for any line L, M + n? If so, find the value of M + N, otherwise, explain the reason

It is known that the focus F of the parabola C: y2 = 2px (P & gt; 0) coincides with the right focus of the ellipse x24 + Y23 = 1, the line L crosses the parabola at two points a and B through point F, and the projections of points a and B on the directrix of the parabola C are points D and e respectively. (I) the process of finding the parabola C; (II) if the line L crosses the y-axis at point m, and Ma = MAF, MB = NBF, is the fixed value for any line L, M + n? If so, find the value of M + N, otherwise, explain the reason

(I) the equation of the right focus f (1,0), P2 = 1, P = 2, parabola C of an ellipse is y2 = 4x (3 points) (II) it is known that the slope of the straight line l must exist, so let L: y = K (x-1), the intersection of L and Y axis at M (0, - K), the intersection of the straight line L with parabola at a (x1, Y1), B (X2, Y2), and y = K (...)

If a focus of the ellipse x2 / M2 + Y2 / (3-m) = 1 is (0,1), then the value of M is

Because the focus is (0,1), so C = 1, because it is an ellipse with the focus on the Y axis, so we have the equation 3-m = M2 + 1
The solution is m = - 2 or M = 1 verification: because it is an ellipse with the focus on the Y axis, so 3-m > M2 + 1, that is, M = - 2, M = 1 all hold
So m = - 2 or 1

X 2 / 4 + y 2 / 3 = 1, f is the right focus of the ellipse, a (1,1) is a certain point in the ellipse, P is a moving point on the ellipse
Finding the minimum value of PA + PF
Finding the minimum value of PA + 2pF

P(x,y)
F(1,0)
e=c/a=1/2
e=PF/(4-x)
PF / E = 4-x, the distance from P to the right guide line
PA+2PF
=PA+PF/e
Obviously
In a straight line, the shortest = 3

If we know the ellipse and hyperbola in common focus, the focus is F1 and F2, and one of their intersection points is p, and ∠ f1pf2 = 120 °, then the relationship between the eccentricity E1 of the ellipse and the eccentricity E2 of the hyperbola must be ()
A. 14e1+34e2=1B. 34e12 +14e22=1C. 34e12+14e22=1D. 14e12+34e22 ＝1

Let the focal length be 2c, the long axis of the ellipse be 2a, and the real axis of the hyperbola be 2m. Let p be defined by the hyperbola | Pf1 | - | PF2 | = 2m on the right branch of the hyperbola. ① by the definition of the ellipse | Pf1 | + | PF2 | = 2A & nbsp; and ∠ f1pf2 = 1200, so | Pf1 | 2 + | PF2 | 2 + | Pf1 | PF2 | = 4C2 & nbsp; & nbsp; ③ ① 2 + ② 2 can get | Pf1 | 2 + | PF2 | 2 = 2A2 + 2M2 ④ - ① 2 + ② 2 can get | Pf1 | PF2 | 2 = a2-m2 ⑤ substitute ④ ⑤ into ③ to get 3a2 + M2 = 4C2, that is 34 × c2a2 + 14 × c2m2 = 1, that is 34e12 + 14e22 = 1, so select C