It is known that the left and right focus of ellipse C1: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 is F 1, F 2, and the eccentricity is 1 / 2. The parabola C2: y ^ 2 = 4mx (M > 0) has a common focus f 2 (1,0) with ellipse C1 (1) Solving equations of ellipse and parabola (2) Let the line L pass through the left focus F 1 of the ellipse and intersect with the parabola at two different points P Q, and satisfy the vector f 1p = A and the vector f 1q, then the value range of the real number a is obtained

It is known that the left and right focus of ellipse C1: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 is F 1, F 2, and the eccentricity is 1 / 2. The parabola C2: y ^ 2 = 4mx (M > 0) has a common focus f 2 (1,0) with ellipse C1 (1) Solving equations of ellipse and parabola (2) Let the line L pass through the left focus F 1 of the ellipse and intersect with the parabola at two different points P Q, and satisfy the vector f 1p = A and the vector f 1q, then the value range of the real number a is obtained

(1) Ellipse C1: X & # 178 / 4 + Y & # 178 / 3 = 1, parabola C2: Y & # 178; = 4x (2) let the linear equation be y = K (x + 1), and substitute it into the parabola equation to sort out K & # 178; X & # 178; + (2k & # 178; - 4) x + K & # 178; = 0. Because there are two intersections between the line and the parabola, △ = (2k & # 178; - 4) &# 178; - 4K ^ 4 > 0, the solution is as follows

It is known that F1 and F2 are the left and right focuses of the ellipse, and the parabola C takes F1 as the vertex and F2 as the focus. Let p be an intersection of the ellipse and the parabola, and
And [Pf1] = e "PF2", what is the value of E?

If F 1 (- C, 0), F 2 (C, 0), the vertex F 1 and focus F 2 of the parabola, then the Quasilinear x = - 3C. Furthermore, Pf1: the distance from P to the left quasilinear of the ellipse = e = [Pf1]: [PF2], so the distance from P to the left quasilinear of the ellipse = PF2, that is, the left quasilinear of the ellipse is the Quasilinear of the parabola, 3C = A & sup2 / / C, thus e = √ 3 / 3

It is known that the eccentricity of ellipse x2 / A2 + Y2 / B2 = 1 (a > b > 0) is √ 6 / 3, and the right focus is (2 √ 2,0)
The straight line L with slope 1 intersects with ellipse g and points a and B, with ab as the base and P (- 3,2) as the vertex
(1) The equation for finding the ellipse G
(2) Finding the area of △ PAB

(1) C / a = √ 6 / 3, C = 2 √ 2, get a = 2 √ 3, a & # 178; = 12, B & # 178; = 4, then the elliptic equation is X & # 178 / 12 + Y & # 178 / 4 = 1 (2), let the midpoint of AB be d (x0, Y0), the linear AB equation be y = x + B, then the simultaneous equation is 4x & # 178; + 6bx + 3B & # 178; - 12 = 0x1 + x2 = - 3B / 2, Y1 + y2 = X1 + x2 + 2B = B / 2, then x0

It is known that a focus f (1,0) of ellipse C: x2 / A2 + Y2 / B2 = 1 (a > b > 0) and eccentricity is 1 / 2
1. Solve the elliptic equation. 2. If M is the left vertex of the ellipse, a straight line with slope k = 1 passing through the focus f intersects the ellipse at two points a and B, and a straight line am and BM intersect with a straight line x = m (M greater than 2) at two points P and Q respectively, and FP is perpendicular to FQ, the value of M is obtained

Obviously, the focal length and half focal length C = 1, from E = 1 / 2 = C / A; so a = 2, then B ^ 2 = a ^ 2-C ^ 2, get B ^ 2 = 3, then the equation is x ^ 2 / 4 + y ^ 2 / 3 = 1. From the point oblique equation, the equation of straight line AB is y = X-1, and the ellipse return is simultaneous to x1x2 = - 8 / 7; X1 + x2 = 8 / 7. Let a coordinate be (x1, x1-1) and B coordinate be (X2, X2