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Is the focal length of ellipse C or 2C
The focal length of the ellipse is 2C. (1) when the focus is on the x-axis, the standard equation is: X & # / A & # / 178; + Y & # / B & # / 178; = 1 (a > b > 0) (2) when the focus is on the y-axis, the standard equation is: Y & # / A & # / 178; + X & # / B & # / 178; = 1 (a > b > 0). The sum of the distances from any point on the ellipse to F1, F2 is 2a, F1, F2
2C
The focal length of the ellipse is C. if the abscissa of the line y = 2C and a focus of the ellipse is exactly C, then the eccentricity of the ellipse is C_ Detailed steps
The coordinates of intersection (C, 2C) are also known
So C ^ 2 / A ^ 2 + 4C ^ 2 / b ^ 2 = 1
B ^ 2 = a ^ 2-C ^ 2
So C ^ 2 / A ^ 2 + (4C ^ 2 / A ^ 2) / (1-C ^ 2 / A ^ 2) = 1
e^2+4e^2/1-e^2=1
So the root number is 2-1
It is known that the focal length of the ellipse x2a2 + y2b2 = 1 (a > b > 0) is 2c, and a, B, C form an arithmetic sequence in turn, then the eccentricity of the ellipse is ⊙___ .
∵ a, B, C in turn into a sequence of arithmetic numbers, ∵ 2B = a + C, A2-B2 = C2, ∵ a2 - (a + C2) 2 = C2, namely & nbsp; 3a2-5c2-2ac = 0, ∵ 5e2-2e + 3 = 0, e = 35 & nbsp; or & nbsp; e = - 1 (rounding off)
Given that the center of the hyperbola is at the origin, the focus is on the y-axis, the focal length is 16, and the eccentricity is 43, the equation of the hyperbola is______ .
Let the hyperbolic equation y2a2 − x2b2 = 1 (a > 0, b > 0), then the focal length of ∵ hyperbola is 16, eccentricity is 43, ∵ 2C = 16ca = 43, ∵ C = 8, a = 6, ∵ B2 = c2-a2 = 28 ∵ hyperbolic equation y236 − x228 = 1, so the answer is: y236 − x228 = 1
It is known that sin (PAI / 4-x) = 3 / 5 and 17 / 12 Pai
The range of π / 4 + X is 5 π / 3
The analytic expression of the inverse scale function is obtained from the image of known inverse scale function through points (1,3). (2) points a (- 1, - 3) and B (- 3, - 1) are on this function image
(1) Let the analytic expression of inverse scale function be y = x / K
(1,3) substituting k = xy = 3, so: the analytic expression of inverse scale function is y = 3 / X
(2) Because: when x = - 1, y = - 3
When x = - 3, y = - 1
So: points a and B are on the graph of this function
(1) Let the inverse scale function be y = K / x, because its image passes through points (1,3),
So 3 = K / 1, k = 3, then the analytic expression of inverse scale function y = 3 / X.
(2) When x = - 1, y = 3 / (- 1) = - 3, so point a (- 1, - 3) is on the function graph;
When x = - 3, y = 3 / (- 3) = - 3, so point B (- 3, - 1) is on the function graph;
Y = 3 / x, substituting x = - 1 into y = 3 / x, y = - 3
And substituting x = - 3 into y = - 1
So point a. B on inverse scale function
The focal point is on the y-axis, the focal length is 20, and the asymptote equation is a hyperbolic standard equation with x = plus or minus 4 / 3Y
Standard equation: Y2 / 64-x2 / 36 = 1 (Y2, X2 is their square)
The focal length is 20, C = 10. According to the asymptote, we can know the ratio between a and B. because the focus is on the Y axis, a: B = 4:3
According to C2 = A2 + B2, we can get a = 8, B = 6, that is, A2 = 64, B2 = 36, and then we can get the answer
Given sin (a-pai) = 3 / 5, find the value of COS 2A
sin(a-π)=3/5,
-sin(π-a)=3/5
sina=-3/5
cos2a=1-2sin²a=1-2×(9/25)=7/25
The final result needs a prescription
If the image of an inverse scale function passes through a point (- 2,3), then the function image passes through a point ()
A. （2，-3）B. （-3，-3）C. （2，3）D. （-4，6）
Let the inverse scale function y = KX (k is a constant, K ≠ 0), ∵ the image of the inverse scale function passes through points (- 2,3), ∵ k = - 2 × 3 = - 6, and 2 × (- 3) = - 6, (- 3) × (- 3) = 9, 2 × 3 = 6, - 4 × 6 = - 24, ∵ points (2, - 3) are on the image of the inverse scale function y = - 6x
The standard equation of hyperbola with real axis 6, focal length 8, focus on X axis and center at origin?
The focus is on the X axis
So real axis length = 2A = 6
A=3
Imaginary axis length = 2B = 8
B=4
The focus is on the X axis and the center is at the origin
So x & sup2; / A & sup2; - Y & sup2; / B & sup2; = 1
So x & sup2 / 9-y & sup2 / 16 = 1
The focal length of the first floor seems to be 2C??
It should be 2C
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