![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
The focus of ellipse X / 5A + Y / (4a + 1) = 1 is on the x-axis, and the value range of eccentricity is calculated
The focus of the ellipse X / 5A + Y / (4a + 1) = 1 is on the x-axis, satisfying
5A > 4A ＾ 2 + 1
1/4
If the focus of ellipse (X & sup2 / 5A) + [y & sup2; / (4a & sup2; + 1)] = 1 is on the x-axis, then the value range of its eccentricity is_____ (reason?)
5A＞4A²+1
(A-1)(4A-1)＜0
0.25＜A＜1
e=c/a
e²=1-b²/a²=1-（4A²+1）/5A
(4a & sup2; + 1) / 5A ≥ 4 / 5, if and only if a = 1 / 2
When a = 1, (4a & sup2; + 1) / 5A = 1
When a = 1 / 4, (4a & sup2; + 1) / 5A = 1
∴0＜e≤√5/5
That's right
RELATED INFORMATIONS
- 1. The ellipse with focus on X axis X ^ 2 / 4A + y ^ 2 / (a ^ 2 + 1) = 1, and the maximum eccentricity is
- 2. Find a point P on the ellipse x ^ 2 / 20 + y ^ 2 / 56 = 1, so that the line between point P and two focal points is perpendicular to each other
- 3. The coordinates of the point on the parabola y2 = 4x with the closest distance to its focus are
- 4. If the sum of the distances from a point a to B (3,2) on the parabola y2 = 4x to the focus is the smallest, then the coordinate of point a is______ ．
- 5. Given that the parabola x 2 = 4Y, point P is a moving point on the parabola, and the coordinates of point a are (12,6), the minimum value of the sum of the distance from point P to point a and the distance from point P to X axis is obtained
- 6. Given that the distance difference between the moving point P and the points F1 (- 5,0) and F2 (5,0) is 6, then the trajectory equation of point P is?
- 7. Given two fixed points F1 (- 5,0), F2 (5,0), find the trajectory equation of point P whose absolute value of the difference between the distances of F1 and F2 is 6
- 8. The trajectory equation of the point on the plane where the absolute value of the distance difference between two fixed points F1 (- 7,0) F2 (7,0) is equal to 10 2c=14 c=7 c^2=49 2a=10 a=5 a^2=25 b^2=49-25=24 x^2/25+y^2/24=1 But the answer is x ^ 2 / 25-y ^ 2 / 24 = 1 Shouldn't the + sign be used in the middle? Is the trajectory equation + or -?
- 9. Let F1 and F2 be fixed points, | F1F2 | = 6, and the moving point m satisfy | MF1 | + | MF2 | = 6, then the trajectory of the moving point m is () A. Ellipse B. straight line C. circle D. line segment
- 10. There are two fixed points F1 (0, - 4) (0, - 4) in the plane, and the moving point satisfies the trajectory equation of mf1-mf2 absolute value = 8, M
- 11. 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
- 12. Given the eccentricity of ellipse C E = √ 6 / 3, a Quasilinear equation is x = 3 √ 2 / 2 Let P satisfy: OP vector = om vector + on vector, where m and N are the points on the ellipse, and the product of the slopes of OM and on is - 1 / 3. Question: are there two fixed points a and B, so that PA + Pb is the fixed value? If so, find the coordinates of a and B; if not, explain the reason
- 13. It is known that the center of ellipse C is the origin of coordinate, the length of minor axis is 2, and a Quasilinear equation is l: x = 2
- 14. If a line passing through a focus F1 of ellipse 4x2 + 2Y2 = 1 intersects with ellipse at two points a and B, then a and B and another focus F2 of ellipse form △ abf2, then the perimeter of △ abf2 is () A. 2B. 22C. 2D. 1
- 15. The focus coordinates of the ellipse X25 + y24 = 1 are______ ．
- 16. The chord passing through the focus of the parabola and perpendicular to the axis of symmetry is called the path of the parabola. Find the equation of the parabola whose vertex is at the origin and the path length is 8 Because the topic didn't give the focus position, it was discussed in four situations . How do we get 2p = 8?
- 17. (2011 · Anhui simulation) the area of △ pf1f2 is () A. 20B. 22C. 24D. 28
- 18. From an end point of the ellipse minor axis to see the two focal points, the angle of view is 120 to calculate the eccentricity
- 19. The two ends of the minor axis of the ellipse form 120 degrees with the line connecting the focus, and the eccentricity of the ellipse is calculated
- 20. The center of a circle is on the right focus of the ellipse 16x ^ 2 + 25y ^ 2 = 400, and passes through the vertex of the ellipse on the y-axis to find the equation of the circle