If the vertex of the minor axis and the two focal points of the ellipse form an equilateral triangle, then the C / a of the ellipse is equal to?

If the vertex of the minor axis and the two focal points of the ellipse form an equilateral triangle, then the C / a of the ellipse is equal to?

Since the distance from the vertex of the minor axis of the ellipse to the focus is a,
An equilateral triangle is composed of a vertex of the minor axis and two focal points
a=2c
Then C / a = 1 / 2

If a vertex and two focal points of an ellipse form an equilateral triangle, then the eccentricity of the ellipse is?

Side length a = 2C
Eccentricity e = 1 / 2

It is known that the left vertex of the ellipse is a, and two perpendicular chords am and an intersect the ellipse at two points m and N through A. (1) when the slope of the line am is 1, calculate the coordinates of point m; (2) when the slope of the line am changes, whether the line Mn passes a certain point on the x-axis. If it passes a fixed point, please give a proof and find the fixed point. If it fails, please explain the reason

(1) When the slope of the straight line am is 1, the straight line am: y = x + 2, (1 point) is substituted into the elliptic equation and simplified to be: 5x2 + 16x + 12 = 0, (2 points) the solution is X1 = - 2, X2 = - 65, ｜ M (- 65, 45); (4 points) (2) let the slope of the straight line am be K, then am: y = K (x + 2), then y = K (x + 2) x24 + y2 = 1, and the simplification is as follows:

It is known that the left vertex of the ellipse x 24 + y 2 = 1 is a. two perpendicular chords am and an intersect the ellipse at M and N through a
Known ellipse
The left vertex of x 2 / 2 + y 2 / 2 = 1 is a. through a, two perpendicular chords am and an intersect ellipses at M and n
When the slope of line am is 1, the coordinates of point m and the intersection coordinates of line Mn and X-axis are obtained

Am equation can be solved, y = x + 2, and m point coordinate can be obtained by combining with ellipse
An ⊥ am, the linear an equation is y = - X - 2, and the coordinate of n points can be obtained by combining with ellipse
The linear Mn equation can be solved, so that y = 0 and X-axis intersection coordinates

The left vertex of x2 / 4 + Y2 / 2 = 1 is a. two perpendicular chords am and an intersect ellipses are made through a at two points m and n
The left vertex of x 2 / 4 + y 2 / 2 = 1 is a. through a, two perpendicular chords am and an intersect ellipses at M and n
When the slope of line am is 1, the coordinates of point m and the intersection coordinates of line Mn and X-axis are obtained

If the a coordinate is known and the AM slope is known, the AM equation, the simultaneous am equation and the elliptic equation can be written out, and the m-point coordinate can be obtained. The m-point coordinate can be obtained quickly by using Veda
Similarly, since am and an are perpendicular, the slope of an is known to be - 1. Write an equation, and combine it with the elliptic equation to solve the N coordinate,
M. When n coordinates are solved, Mn equation is known, and the intersection point with X axis is easy to be solved

Go through the right vertex a of the ellipse C: x ^ 2 / 4 + y ^ 2 = 1, and make two mutually perpendicular lines am and an to intersect the ellipse C at two points m and N respectively
If the slope of AM line is k, the coordinates of point m are obtained

It can be seen from the problem that the coordinate of a is (2,0), and the two sides are multiplied by 4 to get x ^ 2 + 4Y ^ 2 = 4. Let am be y = KX and substitute it into the above formula to get (4K ^ 2 + 1) x ^ 2 = 4, that is, x = positive and negative root sign 4 / (4K ^ 2 + 1). So m (positive and negative root sign 4 / (4K ^ 2 + 1), K times positive and negative root sign 4 / (4K ^ 2 + 1)). K exists and is not equal to 0

The center of the ellipse is at the origin, a vertex is (2,0) and the length of the minor axis is equal to the focal length

Because B = C, 1) if the vertex (2,0) is the endpoint of the minor axis of the ellipse, then B = C = 2, so a ^ 2 = B ^ 2 + C ^ 2 = 8, so the standard equation of the ellipse is y ^ 2 / 8 + x ^ 2 / 4 = 1.2) if the vertex (2,0) is the endpoint of the major axis of the ellipse, then a = 2, so B ^ 2 = (b ^ 2 + C ^ 2) / 2 = a ^ 2 / 2 = 2, so the standard equation of the ellipse is x

Let the right focus of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a ＞ B ＞ 0) be f and the right guide line be L. if the chord length passing through the point F and perpendicular to the X axis is equal to that from the right vertex to the right guide line
The eccentricity of the ellipse is 2 times the distance L

Substituting x = C into the elliptic equation to get 2|y |, the chord length of point F and perpendicular to X axis is 2B ^ 2 / A, the distance from right vertex to right quasi-l is a ^ 2 / C-A, so 2B ^ 2 / a = 2A ^ 2 / C-2A, so B ^ 2 / a = a ^ 2 / C-A, so B ^ 2 / A ^ 2 = A / C-1, that is, 1-e ^ 2 = 1 / E-1 = (1-e) / E, so 1 + e = 1 / E, that is, e ^ 2 + E-1 = 0, e = (- 1 + √ 5) / 2

If the focal length of the ellipse is equal to the distance between one end of the major axis and one end of the minor axis,

According to the meaning of the title, the length of the line segment between the two endpoints is the root sign (a ^ 2 + B ^ 2)
Column equation a ^ 2 + B ^ 2 = (2C) ^ 2
a^2-b^2=c^2
Two formula addition
2a^2=5c^2
E = C / a = radical 0.4

The focal length of an ellipse is equal to the distance between an end point of the major axis and an end point of the minor axis, and the eccentricity of the ellipse is calculated

According to the meaning of the title: √ A2 + B2 = 2c, that is, A2 + B2 = 4C2
And because B2 = a2-c2
So 2A2 = 5c2
c2/a2=2/5
Eccentricity of ellipse e = C / a = √ 10 / 5