As shown in the figure, F1 and F2 are the left and right focal points of the ellipse respectively. The abscissa of the point m on the ellipse is equal to the abscissa of the right focal point, and its ordinate is equal to 23% of the length of the short half axis. Calculate the eccentricity of the ellipse

As shown in the figure, F1 and F2 are the left and right focal points of the ellipse respectively. The abscissa of the point m on the ellipse is equal to the abscissa of the right focal point, and its ordinate is equal to 23% of the length of the short half axis. Calculate the eccentricity of the ellipse

The focus is F1 (- C, 0), F2 (C, 0), and the coordinate of point m is (C, 23B) (C, 23B) (C, 23B), and the coordinate of point m is (C, 23B) (r t \\\\\\\\\\\\\|f1f2 |2 + _|mf2-2|2 |2 | MF2 |2 = (2a-mf2) 2 = (2a) MF2 = (2a) 2 = (2a-mf2) 2 = (2a-mf2) 2 = (2a-mf2) 2 (2) 2) 2 = (2a-2) 2) 2 = (2a-mf2) 2-2 = (2a-2(2a-23b) 2 = 4C2 + 49b2 If we get 4C2 = 4a2-83ab, we can get 3 (a2-c2) = 2Ab, so 3B2 = 2Ab, the solution is b = 23a, ∧ C = A2 − B2 = 53a, so we can get e = CA = 53, that is, the eccentricity of the ellipse is equal to 53

F1 and F2 are the left and right focal points of the ellipse respectively. The abscissa of the point m on the ellipse is equal to the abscissa of the right focal point, and its ordinate is equal to the abscissa of the short half axis
⅔, find the eccentricity of ellipse,

This test is the knowledge point of path: in an ellipse, the chord passing through the focus perpendicular to the major axis is called path, its length d = 2B ^ 2 / A, so the half path = B ^ 2 / A; the abscissa of point m is equal to the abscissa of the right focus, which means that point m is the intersection of path and ellipse, so its ordinate = ± B ^ 2 / A; from the title: B ^ 2 / a = 2B / 3, get: B / a = 2 / 3

Let F 1 and F 2 be the left and right focal points of the ellipse x 24 + y 23 = 1, and make a straight line passing through the center of the ellipse to intersect with the ellipse at P and Q. when the area of the quadrilateral pf1qf 2 is the largest, the value of Pf1 · PF2 is equal to ()
A. 0B. 1C. 2D. 4

Because the ellipse equation is x24 + Y23 = 1, so a = 2, B = 3, so C = A2 − B2 = 1. When the area of the quadrilateral pf1qf2 is the largest, the points P and Q are just the ends of the minor axis of the ellipse. At this time, Pf1 = PF2 = a = 2. Because the focal length | F1F2 | = 2C = 2, so △ pf1f2 is an equilateral triangle, so ∠ f1pf2 = 60 °, so Pf1 · PF2 = 2 × 2 × cos60 ° = 2, so choose C

Let a and f be the left vertex and right focus of the ellipse x ^ 2 / A ^ 2 + y ^ 2 + B ^ 2 = 1 (a > b > 0), respectively. If there is a point P on the right guide line, the vertical plane of the line Pa will be obtained
If the vertical bisector of the line PA just passes through the point F, the range of eccentricity of the ellipse is?

According to the meaning of the title, PF = AF = a + C ≥ d = A & sup2 / C-C (D is the distance between F and the guide line)
a+c≥(a²-c²)/c
ac+c²≥a²-c²
And then divide by a & sup2;, we can get
c/a + c²/a²≥1-c²/a²
e+e²≥1-e²
2e²+e-1≥0
The solution is 0.5 ≤ e < 1
When we take the equal sign, AF = D, the vertical line must pass through F

It is known that the left vertex of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 is a, the upper vertex is B, and the right focus is F. let the midpoint of AB be m, and if 2mA * MF + BF ^ 2 > 0, then the eccentricity of the ellipse
Given that the left vertex of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 is a, the upper vertex is B, and the right focus is f, let the midpoint of AB be m. if the 2 vector ma * vector MF + vector BF ^ 2 > = 0, then the eccentricity of the ellipse is?

Find a point P on the ellipse 9x ^ 2 + 25y ^ 2 = 225 so that its distance from the left focus is twice that from the right focus
Urgent!

x^2/25+y^2/9=1
a=5
So Pf1 + PF2 = 2A = 10
Let F1 be the left focus
So Pf1 = 2pf2
So 2pf2 + PF2 = 10
PF2=10/3
c^2=a^2-b^2=16
So F2 (4,0)
P(m,n)
So PF2 ^ 2 = (M-4) ^ 2 + n ^ 2 = (10 / 3) ^ 2
P is on the ellipse
So m ^ 2 / 25 + n ^ 2 / 9 = 1
n^2=9-9m^2/25
So (M-4) ^ 4 + 9-9m ^ 2 / 25 = 100 / 9
9m^2-72m+125-81m^2/25=0
144m^2-1800m+3125=0
(12m-125)(12m-25)=0
m=125/12,m=25/12
Because a = 5, the maximum abscissa is 5
So m = 25 / 12, n ^ 2 = 9-9m ^ 2 / 25 = 119 / 16
So p (25 / 12, √ 119 / 4) or P (25 / 12, √ 119 / 4)

It is known that the eccentricity of the ellipse X & # 178 / / A & # 178; + Y & # 178 / / B & # 178; = 1 (a ＞ B ＞ 0) is √ 2 / 2, the intersection ellipse of the straight line passing through point B (0, - 2) and left focus F1 is at two points c and D, and the right focus is set as F2.1. The equation for solving the ellipse. 2, the area of △ CdF2

(1) ∵ one vertex of ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is a (0,1),
The centrifugation is √ 2 / 2,
And C / a = √ 2 / 2,
The solution is a = √ 2, C = 1
The equation of the ellipse is
x^2/2 +y^2＝1；
(2) ∵ left focus F1 (- 1,0), B (0, - 2), the slope of F1B line is - 2
The equation of line F1B is y = - 2x-2
By y = (?) 8722; 2x (?) 8722; 2
x^2/2+y^2=1,
It is reduced to 9x ^ 2 + 16x + 6 = 0
∵△=162-4×9×6=40＞0,
The line and the ellipse have two common points, set C (x1, Y1), D (X2, Y2),
be
x1+x2＝−16/9
x1•x2＝2/3
∴|CD|=√1+(−2)^2|x1-x2|=√5 *√[(x1+x2)^2-4x1x2]
=√5 *√[(-16/9)^2-4*2/3]=10√2/9
And the distance between point F2 and straight line BF1 d = |; 2 &; 2 | / √ 5 = 4 √ 5 / 5,
The area of | CdF2 is s = 1 / 2 | CD | × D
=1/2 * 10√2/9 * 4√5/5
=4√10/9

It is known that the ellipse C: X & # 178 / / A & # 178; + Y & # 178 / / B & # 178; = 1 (a ﹥ B ﹥ 0) f is its focus, and the eccentricity is e
(1) If the Quasilinear of the parabola x = 1 / 8y & # passes through point F and the ellipse C passes through P (2,3), the equation of the ellipse C is obtained
(2) If the line of a (0, a) is tangent to the ellipse C at M and intersects the X axis at B, and the vector am = μ, the vector Ba proves μ + E & # 178; = 0

(1) The X-ray is x = -2, then x = -2 passes through the left focus of the ellipse F1, so C = 2 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\17

It is known that the right focus f (1,0) of ellipse C: X & # 178; + Y & # 178; = 1 (a > b > 0), the eccentricity is 1 / 2, the line L passing through point F intersects ellipse C at two points a and B, and 27 / 11 ≤| FA | ×| FB | ≤ 3

If the intersection of a straight line passing through a focus F of the ellipse 4x & # 178; + Y & # 178; = 1 and two points of the ellipse A and B, F & # 8322; is another focus of the ellipse, then the triangle
What is the circumference of ABF?

x²/(1/4)+y²/1=1
1/4