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Taking the focus of the ellipse as the center of the circle and the focal length as the radius of the circle passing through the two vertices of the ellipse to calculate the eccentricity of the ellipse~- Taking the focus of the ellipse as the center of the circle and the focal length as the radius of the circle passing through the two vertices of the ellipse to calculate the eccentricity of the ellipse~ -In addition, do you need to consider which axis the ellipse is on
The final result is the same without the focus on the X and Y axes
Let the ellipse focus on the x-axis, the left vertex A1 (- A, 0), the right vertex A2 (a, 0), the lower vertex B1 (0, - b), and the upper vertex B2 (0, b)
Left focus F1 (- C, 0), right focus F2 (C, 0)
From the title: f1b2 = F1F2,
That is, a = 2C
So e = C / a = 1 / 2
The left and right focal points of hyperbola are F1 and F2. Point P is on the right branch of hyperbola, and Pf1 = 4pf2. Find the maximum value of eccentricity e of hyperbola
Hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1
∵|PF1|=4|PF2|
P is on the right branch,
∵ according to the definition of hyperbola, | Pf1 | - | PF2 | = 2A
∴4|PF2|-|PF2|=2a
∴|PF2|=2/3*a
∵ the distance between P and F2 on the right branch of hyperbola: | PF2 | is [C-A, + ∞)
∴2/3*a≥c-a
∴c≤5/3a
∴e=c/a≤5/3
And E > 1
∴1
If the hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1, F1 and F2 are the left and right focal points, and P on the right branch satisfies | Pf1 | = 4 | PF2 |, then the maximum eccentricity of the curve is
Given that the hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1, F1 and F2 are the left and right focal points, and a point P on the right branch satisfies | Pf1 | = 4 | PF2 |, then the maximum value of hyperbolic eccentricity is | Pf1 | = 4 | PF2 |?
Let P (x, y) Pf1 = m PF2 = n
M＝ex＋a N＝ex－a
M = 4N
E = 5AX / 3, when x = B, X is the largest
Therefore, Emax = 5ab / 3
RELATED INFORMATIONS
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