It is known that the hyperbolic equation is x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 (a > O, b > o), and the distance from a focus of hyperbola to an asymptote is √ 5C / 3 to calculate the eccentricity The asymptote equation of hyperbola is y = (± B / a) x, and an asymptote is y = (B / a) x or BX ay = 0 A focus of hyperbola f (C, 0), a ^ 2 + B ^ 2 = C ^ 2 The distance from focus f to an asymptote is: |bc-a*0|/c=b b=√5c/3,9b^2=5c^2 From: A ^ 2 + B ^ 2 = C ^ 2 I know the answer to this question, and then I want to ask why a ^ 2 = 4 / 9C ^ 2

9b²=5c²
9(c²-a²)=5c²
4c²=9a²

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1. Hyperbola has the same focus as ellipse x square + 2Y square = 20, its asymptote is 3x-y = 0, then hyperbolic equation is?
2. The left and right focal points of hyperbola x2 / a2-y2 / B2 = 1 (a > 0, b > 0) are F1 and F2 respectively. The segment F1F2 is divided into two segments of 3:2 by the point (B / 2,0), which is the eccentricity of hyperbola
3. Let the hyperbola x 2 / 4-y 2 / 9 = 1, f 1F 2 be the two focal points. The point m is on the hyperbola (1) If ∠ f1mf2 = 90 °, calculate the area of △ f1mf2 (2) If ∠ f1mf2 = 60 °, what is the area of △ f1mf2? If ∠ f1mf2 = 120 °, what is the area of △ f1mf2?
4. Given that the right focus of the hyperbola x2a2-y25 = 1 is (3,0), then the eccentricity of the hyperbola is equal to () A. 31414B. 324C. 32D. 43
5. It is known that F1F2 is the left and right focus of hyperbola x2 / a2-y2 / B2 = 1 (a > 0, b > 0), the line passing through point F1 and perpendicular to the real axis and the two asymptotes of hyperbola If the coordinate origin o is just the vertical center of △ abf2 (the intersection of the three high lines of the triangle), then the eccentricity of the hyperbola is
6. Let the eccentricity of hyperbola be √ 5 / 2 and have a common focus with the square of ellipse x2 / 9 + the square of Y2 / 4 = 1
7. The standard hyperbolic equation with focus on the x-axis, imaginary axis length of 12 and eccentricity of 54 is () A. x264−y2144＝1B. x236−y264＝1C. y264−x216＝1D. x264−y236＝1
8. Given that the eccentricity of hyperbola is 2, the focus is (4,0), (- 4,0), then what is the hyperbolic equation? Help me if you know
9. It is known that the right quasilinear of two hyperbolas is x = 4, the right focus is f (10.0), and the eccentricity is e = 2?
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11. If x / A + Y / b = 1, the distance from the abscissa A / 3 point to the left focus is greater than the distance from the abscissa A / 3 point to the right guide line, then the eccentricity range of the ellipse is larger
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13. If there is a point P on an ellipse or hyperbola and the distance ratio between the point P and the two focal points is 2:1, then the eccentricity of the ellipse is () A. [14，13]B. [13，12]C. (13，1)D. [13，1)
14. If there is a point P on an ellipse or hyperbola and the distance ratio between the point P and the two focal points is 2:1, then the eccentricity of the ellipse is () A. [14，13]B. [13，12]C. (13，1)D. [13，1)
15. It is known that the focus of hyperbola is F1 (- 6.0), F2 (6.0), and it passes through the point P (- 5.0), so the hyperbolic standard equation is solved
16. It is known that the right focus of the hyperbola x2 / 9-y2 / 16 = 1 is F1, F2, the point P is on the left point of the hyperbola, and the absolute value of Pf1 multiplied by the absolute value of PF2 is equal to 32, so the size of the angle f1pf2 is calculated
17. If the right focus of hyperbola x ^ 2 / 9-y ^ 2 / 16 = 1 is F1, point a (9,2), and point m is on the hyperbola, then the distance from the minimum value m of Ma + 3 / 5mf1 to F1 is greater than that from it to the right "The distance between M and F1 is greater than the distance between M and the right guide line, d = e = C / a = 5 / 3 d=3/5MF1 Ma + 3 / 5mf1 = ma + d > = m distance to right guide line = 9-9 / 5 = 36 / 5 " Why is Ma + d not directly equal to the abscissa of a: 9?
18. The two focal coordinates of the hyperbola are (- 5,0) (5,0), and the absolute value of the distance difference between the two focal points on the hyperbola is 8
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