It is known that the right focus a of ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is the focus of parabola y ^ 2 = 8x, the vertex on the circle is B, and the eccentricity is √ 3 / 2,1, Find the path of ellipse C; 2, the line L passing through point (0, √ 2) with slope k intersects ellipse C at two points P and Q. if the abscissa of the midpoint of line PQ is 4 √ 2 / 5, find the equation of line L

It is known that the right focus a of ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is the focus of parabola y ^ 2 = 8x, the vertex on the circle is B, and the eccentricity is √ 3 / 2,1, Find the path of ellipse C; 2, the line L passing through point (0, √ 2) with slope k intersects ellipse C at two points P and Q. if the abscissa of the midpoint of line PQ is 4 √ 2 / 5, find the equation of line L

The vertex of the circle is B, which seems a little unclear

It is known that a B is the left and right vertices of the ellipse x ^ 2 / 4 + y ^ 2 / 3 = 1, and F is the right focus of the ellipse,
P is a straight line at any point on the ellipse which is different from a and B. AP and BP intersect the straight line L: x = m (M > 2) respectively
At M N, l intersects X axis at C
Find the minimum value of △ MFN area for any given m value
Senior three mathematics quality inspection of the finale, I hope someone can answer
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Everyone has a different answer? Who is right?

From the symmetry of the graph, let p be in the upper half ellipse. Let P coordinate be (x, y)
Through P as pH ⊥ AB at point H
So pH = y, ha = x + 2, Hb = 2-x, AC = m + 2, BC = m-2
MC/PH=AC/AH
So: MC = pH * AC / ah = y (M + 2) / (x + 2)
NC/PH=BC/BH
So: NC = pH * BC / BH = y (m-2) / (2-x)
MC*NC=y^2(m^2-4)/(4-x^2)
The point P is on the ellipse, so: 3x ^ 2 + 4Y ^ 2 = 12,4-x ^ 2 = 4Y ^ 2 / 3
MC*NC=y^2(m^2-4)/(4y^2/3)=3(m^2-4)/4
This value has nothing to do with the position of P. when m is a fixed value, it is also a fixed value
According to the mean inequality:
MN=MC+NC>=2√MC*NC=√[3(m^2-4)]
That is, the minimum value of Mn is √ [3 (m ^ 2-4)]
FC = M-1 is also a fixed value
Therefore, the minimum area of △ MFN is (m-1) * √ [3 (m ^ 2-4)] / 2
If and only if MC = CN
That is: Y (M + 2) / (x + 2) = y (m-2) / (2-x)
(m+2)/(x+2)=(m-2)/(2-x)=2m/4=m/2
x=(m-2)/2
That is, when x = (m-2) / 2, △ MFN takes the minimum area (m-1) * √ [3 (m ^ 2-4)] / 2

The two vertices a (- 1,0), B (1,0) of the ellipse, the line L passing through its focus f (0,1) intersects the ellipse at two points CD, and intersects with the X axis at point P
The two vertices a (- 1,0), B (1,0) of the ellipse, the line L passing through its focus f (0,1) intersects the ellipse at two points CD, and intersects the X axis with the point P. when the absolute value CD is equal to two-thirds of the root sign 2, the equation of the line L is obtained

In this paper, we use the standard elliptic equation a (- 1,0), B (1,0) = > A ^ 2 = 1, focus f (0,1) = > b ^ 2-A ^ 2 = 1 = > b ^ 2 = 2 elliptic equation: x ^ 2 + y ^ 2 / 2 = 1, straight line L passes (0,1). We can set the equation of straight line L as (Y-1) = KX = > y = KX + 1 and substitute it into the elliptic equation x ^ 2 + (KX + 1) ^ 2 / 2 = 1 = > (1 + K ^ 2 / 2) x ^ 2 + kx-1

With the focus of x24 − y212 = − 1 as the vertex, the elliptic equation with the vertex as the focus is ()
A. x216+y212＝1B. x212+y216＝1C. x216+y24＝1D. x24+y216＝1

Hyperbola x24 − y212 = − 1 has vertices of (0, - 23) and (0, 23), focal points of (0, - 4) and (0, 4).. the focal coordinates of ellipse are (0, - 23) and (0, 23), and the vertices of (0, - 4) and (0, 4).. the elliptic equation is x24 + y216 = 1