Hyperbola focuses on the focus of ellipse X / 9 + Y / 25 = 1, and its eccentricity is twice that of ellipse

Hyperbola focuses on the focus of ellipse X / 9 + Y / 25 = 1, and its eccentricity is twice that of ellipse

X squared / 25-y squared / 39 = 1
How to calculate the eccentricity of ellipse and hyperbola
The definition of eccentricity of circle, ellipse, parabola and hyperbola is e = C / A
Because a > 0, C > = 0, e is nonnegative
It is a circle when e = 0
When 0
The eccentricity of ellipse and hyperbola is e = C / A
① Eccentricity of ellipse and hyperbola I: e = C / A
② The second method of calculating eccentricity of ellipse and hyperbola (also called the second definition of ellipse and hyperbola)
E = distance from ellipse (or hyperbola) to focus / distance from "corresponding" guide line
We need to analyze the specific problems. We can see some examples.
What is the eccentricity of conic? What is the relationship between the eccentricity of ellipse and hyperbola and its shape?
The eccentricity of an ellipse is a measure of how flat it is
The unified definition of eccentricity is the ratio of the distance from the moving point to the focus and the distance from the moving point to the directrix, which is a measure of the flattening degree of the ellipse. Eccentricity is defined as the ratio of the distance between the two focuses of the ellipse and the length of the major axis, expressed by E, that is, e = C / a (C, half focal length; a, The eccentricity of the ellipse can be understood as the degree of the two focal points leaving the center on the premise that the long axis of the ellipse remains unchanged. Eccentricity = (RA RP) / (RA + RP), RA is the distance from the far point, RP is the distance from the near point. Eccentricity of circle = 0 eccentricity of ellipse: e = C / a (... Expansion)
The unified definition of eccentricity is the ratio of the distance from the moving point to the focus and the distance from the moving point to the directrix, which is a measure of the flattening degree of the ellipse. Eccentricity is defined as the ratio of the distance between the two focuses of the ellipse and the length of the major axis, expressed by E, that is, e = C / a (C, half focal length; a, The eccentricity of the ellipse can be understood as the degree of the two focal points leaving the center on the premise that the long axis of the ellipse remains unchanged. Eccentricity = (RA RP) / (RA + RP), RA is the distance from the far point, RP is the distance from the near point. Eccentricity of circle = 0 eccentricity of ellipse: e = C / a (0,1) (C, half focal length; a, eccentricity of semi major axis (ellipse) / semi real axis (hyperbola)) parabola: e = 1 eccentricity of hyperbola: e = C / a (1, + ∞) (c, half focal length; a, Semi major axis (ellipse) / semi real axis (hyperbola)) in the unified definition of conic, the unified polar coordinate equation of conic (quadratic non-circular curve) is ρ = EP / (1-e × cos θ), where e is the eccentricity and P is the distance from the focus to the Quasilinear. The distance from the focal point to the nearest directrix is equal to ex ± a. The relationship between eccentricity and curve shape is as follows: e = 0, circle 0
Given that the pattern of the first-order function y = (M-3) x + 2m-1 passes through quadrants 1, 2 and 4, the value range of M is obtained
Because the pattern of the linear function y = (M-3) x + 2m-1 passes through quadrants 1, 2 and 4, M-3 < 0, m < 3
2m-1 > 0, M > 0.5, so 0.5 < m < 3
If the intersection of the line y = kx-2 (k > 0) and the hyperbola y = K / X in the first quadrant is r, and the intersection of the line y = kx-2 (k > 0) and the X axis is p,
If the intersection of the straight line y = kx-2 (k > 0) and the hyperbola y = K / X in the first quadrant is r, the intersection of the straight line y = kx-2 (k > 0) and the hyperbola y = K / X is p, and the intersection of the straight line y = kx-2 and the hyperbola y = K / X is Q, take RM ⊥ X axis at point m, and if the area ratio of the triangle OPQ and the triangle PRM is 1:1, then the value of K is equal to
∵ RM ⊥ x-axis ∵ RM / / OQ ∵ OPQ is similar to △ MPR ⊥ OP: PM = (1 / 1) ^ (1 / 2) = 1:1 ∵ om: OP = 2:1. It is known that: P (2 / K, 0), m {[1 + (1 + k * k) ^ (1 / 2)] / K, 0} ∵ RM ⊥ x-axis ∵ om = [1 + (1 + k * k) ^ (1 / 2)] / K {[1 + (1 + k * k) ^ (1 / 2)] / K}: (2 / k) =
∵ RM ⊥ X axis,
∴RM//OQ ，
The ∧ OPQ is similar to ∧ MPR,
(∵ the area ratio of triangle OPQ to triangle PRM is 1:1)
The ratio of areas = the square of the ratio of the corresponding edges,
That is: OP: PM = 1:1,
That is: OP = PM
∴OM:OP = 2:1 ，
According to the known conditions, it is not difficult to obtain:
P(2/k ，0) ，
M {[1 + √ (1 + K ^ 2)] / K... Expansion
∵ RM ⊥ X axis,
∴RM//OQ ，
The ∧ OPQ is similar to ∧ MPR,
(∵ the area ratio of triangle OPQ to triangle PRM is 1:1)
The ratio of areas = the square of the ratio of the corresponding edges,
That is: OP: PM = 1:1,
That is: OP = PM
∴OM:OP = 2:1 ，
According to the known conditions, it is not difficult to obtain:
P(2/k ，0) ，
M{[1 + √(1+k^2)]/k ，0}，
∵ RM ⊥ X axis,
∴OM = [1 + √(1+k^2)]/k ，
∴{[1 +√(1+k^2)]/k}:(2/k) = 2:1 ，
х k = 15 / 8 х put away
It is known that f (x) = π / 2
α + β > π is the third quadrant
α-β
It is known that the inverse scale function y = 3-2m / X is in the quadrant where the image is located, and Y decreases with the increase of X. the value range of the letter M is calculated
The image passing through the inverse scale function y = K / X of point (5.1) is as shown in the figure, which is calculated by using the image
(1) When x ≥ 1, the value range of Y
(2) When y ＞ - 2, the value range of X
(PS: 2 coordinates in the picture)
Five point one
One point four
1.3-2m>0,m0,
Y = K / X in quadrant 1 and 3,
When x = 1, y = 5; when y = - 2, x = - 5 / 2,
(1) When x ≥ 1, 0
So the inverse function of Y is less than 0
The derivation is 2m / x ^ 2
If the line y = kx-3 is tangent to the curve y = 2inx, then the real number k=
Because two curves are tangent, the derivative function must have an intersection,
y'=k
y'=2/x
k=2/x,x=2/k
That is, the intersection point is x = 2 / K, where y = - 1, y = 2ln (2 / k)
-1=2ln(2/k)
k=2e^(1/2)
Sin ^ 2 α - cos ^ 2 α = - Cos2 α or sin ^ 2 α times cos ^ 2 α = - Cos2 α. Which one is right and why
(sin squared α minus cos squared α = negative Cos2 α
A:
The front one is right
cos2a=cos²a-sin²a
So: Sin & # 178; a-cos & # 178; a = - cos2a
Given the function y = [(m-1) x ^ m & # 178; - M-1] + m, when m is a value, it is a linear function and the image passes through the 234 th quadrant
Is a function of degree
So M & # 178; - M-1 = 1. Then M = 2 or - 1
"Image passes through 2, 3, 4 quadrants"
So m is less than 0 and M-1 is less than 0, so m = - 1
∵ function y = [(m-1) x ^ m & # 178; - M-1] + m is a linear function and the image passes through the 234 th quadrant,
∴M﹤0，M-1≠0，m²-m-1=1
∴M≠1，M≠1,M1=2，M2=-1
When m = - 1, the function y = [(m-1) x ^ m & # - M-1] + m is a linear function and the image passes through the 234 th quadrant,