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If a straight line passing through the left focus of the ellipse x ^ 2 + 2Y ^ 2 = 4 with a left inclination angle of 30 degrees intersects the ellipse at two points a and B, then the chord length AB = 16 / 5
Linear slope k = tan30 & # 730; = 1 / √ 3
x² + 2y² = 4,x²/4 + y²/2 = 1
c² = a² - b² = 4 - 2 = 2
Left focus f (- √ 2,0)
Linear equation: y = (x + √ 2) / √ 3
Substituting into elliptic equation: 5x & # 178; + 4 √ 2x - 8 = 0
x₁ + x₂ = -4√2/5
x₁x₂ = -8/5
|AB|² = (x₁ - x₂)² + (y₁ - y₂)² = (x₁ - x₂)² + [(x₁ + √2)/√3- (x₂ + √2)/√3)]²
= (x₁ - x₂)² + (x₁ - x₂)²/3
= 4(x₁ - x₂)²/3
= (4/3)[(x₁ + x₂)² - 4x₁x₂]
= (4/3)[(-4√2/5)² - 4(-8/5)]
= 256/25
|AB| = 16/5
The length of the chord ab of the left focus F1 passing through the ellipse x2 / A2 + Y2 / B2 = 1 is 3, af2 = 4, and the vector AB * the vector af2 = 0 is used to calculate the eccentricity
AB * af2 = 0, so AB sum is perpendicular to af2, then by Pythagorean theorem BF2 = 5;
Then according to the first definition of ellipse, AF1 + af2 = BF1 + BF2 = 2A
Because AB + af2 + AF1 = 3 + 4 + 5 = 12, 4A = 12, a = 3;
Then we have AF1 = 2a-af2 = 3, then we have Pythagorean theorem, we have F1F2 = 3, radical 2 = 2C
So C = 3 radical 2 / 2, e = C / a = radical 2 / 2
The minimum perimeter of a triangle formed by a line passing through the origin and a focus in an ellipse
This problem is not completely written. It is speculated that this problem is to find the minimum perimeter of triangle formed by the intersection of line and ellipse and a focus. As shown in the figure, it is easy to find the minimum perimeter of triangle from the properties of ellipse and graph, that is, to find the shortest line segment at the intersection of line and ellipse. When the line coincides with the minor axis, the line segment is the shortest, so l = (2a + 2a) / 2 + 2B = 2 (a + b)
Let p be a point on the ellipse 9x ^ 2 + 25y ^ 2 = 225, and let F1 and F2 be the two focal points of the ellipse. Try to find the minimum and maximum absolute value Pf1 * absolute value PF2
The ellipse is transformed into the standard equation: x ^ 2 / 25 + y ^ 2 / 9 = 1
a^2=25
a=5
2a=10
|PF1|+|PF2|=2a=10
|PF2|=10-|PF1|
|PF1|*|PF2|=|PF1|*(10-|PF1|)
Let | Pf1 | = x (x > = 1 and x)
Given that P is any point on the ellipse x ^ 2 / 9 + y ^ 2 / 3 = 1, F1 and F2 are the two focuses of the ellipse, find the absolute value of Pf1 * the maximum absolute value of PF2?
It's better to have a result
|In this paper, we need to know that PF2 [Pf1 124124\124124124124\\\124\124\124\124\124\\\\124\\\124\\\124\\\\\\\\\\\\\\\\\\\\\\\\\whenpf2 | = a, it satisfies the maximum value (P point is on the Y axis, the endpoint of the minor axis
Let p be a point on the ellipse x ^ 2 / 5 + y ^ 2 / 25 = 1, and F1 and F2 be the two focuses of the ellipse. If Pf1 ⊥ PF2, then the absolute value of the difference between Pf1 and PF2
The focus of the ellipse x ^ 2 / 5 + y ^ 2 / 25 = 1 is on the Y axis, x ^ 2 / A + y ^ 2 / b = 1, so B ^ 2 = 25, a ^ 2 = 5, C ^ 2 = 20
Let Pf1 + PF2 = 2B = 10, | F1F2 | = 2c, let PF2 = m, then Pf1 = 10-m
Pf1 ^ 2 + PF2 ^ 2 = | F1F2 | ^ 2, that is, M1 = 5 + √ 15, M2 = 5 - √ 15
｜PF2｜=5+√15,｜PF1｜=5-√15,|｜PF2｜-｜PF1｜|=2√15.
How to solve m ^ 2 + (10-m) ^ 2 = 80
(10-m）^2=m^2-20m+100
m^2+(10-m)^2=80→m^2-10m+10=0
△=10^2-4*10=60
According to (- B ± √ △) / 2A → (10 ± √ 60) / 2 = 5 ± √ 15
m1=5+√15,m2=5-√15
F1 and F2 are the two focal points of the ellipse x ^ 2 / 25 + y ^ 2 / 9 = 1, AB is the chord passing through F1, | ab | = 8, then | AF1 | + | BF1|=
AB should be the string that AB is passing F2, and | ab | ab | ab | ab | = 8, find | AF1 | + | BF1 | a 124; a | a | BF1 | a | a | a | ab | = 8, find | AF1 | + | BF1 | a | a | a \| | | BF1 | BF1 | BF2 | | | |;;;;;;;; | | | | | | | | BF2 |;;;;;;;;; AF1 | + | BF1 | + 8 = 20 | AF1 | + | BF1 | = 12
Ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) focus F1, F2, a ellipse upper moving point, chord AB, AC respectively through F1, F2, when AC is perpendicular to the X axis, AF1 = 3af2
(1) Find the eccentricity of the ellipse; (2) if the vector AF1 is equal to the vector F1B m times, the vector af2 is equal to the vector F2C n times
Find the value of M + n
(I) let | af2 | = m, then | af2 | = 3M
According to the definition of the problem and ellipse,
(3m)^2-m^2=(2c)^2
3m+m=2a
The elimination of M leads to a ^ 2 = 2C ^ 2
So eccentricity e = root 2 / 2
(II) from (1), B ^ 2 = C ^ 2 = a ^ 2 / 2
So the ellipse is:
x^2+2y^2=2c^2
Let a (x0, Y0) B (x1, Y1) C (X2, Y2)
Then x0 ^ 2 + 2y0 ^ 2 = a ^ 2
A is any point on the ellipse that is different from the endpoint of the major axis,
From the known conditions,
m=-y0/y1
n=-y0/y2
So m + n = - Y0 * (1 / Y1 + 1 / Y2)
The equation of AF1 is
x+c=(x0+c/y0)*y
And because x0 ^ 2 + 2y0 ^ 2 = a ^ 2
Lianlide:
[2y0^2+x0+c]y^2-2cy0*(x0+c)y-c^2y0^2
(3c+2x0)y^2-2y0*(x0+c)y-cy0^2=0
According to Weida's theorem, y0y1 = - cy0 ^ 2 / 3C + 2x0
So. Y1 = - C * Y0 / 3C + 2x0
Similarly, y2 = cy0 / - 3C + 2x0
∴.m+n=-y0*(1/y1+1/y2)=6
In conclusion, when point a is a moving point on the ellipse, it is a fixed value of 6
Ellipse x 2 + a 2 + y 2 = 1 (a ＞ 1), with vertex a of minor axis as right angle vertex, and edges AB and BC intersect ellipse at two points B and C. if the maximum area of triangle ABC is 27 ﹥ 8, find the value of A
If we know that point a is (0,1), move the coordinate system up to point a, and transform the elliptic equation to: x2 + A2 + (Y-1) 2 = 1 (a > 1); if we transform the Cartesian coordinate system to the polar coordinate system: x = rcosa, y = rsina, we can get (rcosa) 2 + A2 + (rsina-1) 2 = 1 (a > 1); if we simplify it, we can get r = 2sina / (cosa2 / A2 + sina2)
If F1 and F2 are the focus of the ellipse x2 / 25 + Y32 / 16 = 1, and P is the point on the ellipse that is not on the x-axis, the trajectory equation of the center of gravity g of the triangle pf1f2 is obtained
From the elliptic equation: a = 5, B = 4, C = 3, the coordinates of F1 and F2 are (- 3,0) and (- 3,0), respectively. Let the coordinates of gravity g be (x, y), and the coordinates of point p be (s, t) (t ≠ 0), then x = (- 3) + 3 + s) / 3, y = (0 + 0 + T) / 3, and the solution is s = 3x, t = 3Y, and Y ≠ 0, and the coordinates of point P satisfy the elliptic equation
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