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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• 1. 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?
• 2. 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
• 3. 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
• 4. 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)
• 5. 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)
• 6. The maximum distance from the moving point P on the standard ellipse with the focus on the x-axis to the top vertex is equal to the distance from the center of the ellipse to its directrix, and the value range of eccentricity is calculated The maximum distance from the moving point P on the standard ellipse with the focus on the x-axis to the top vertex is equal to the distance from the center of the ellipse to its directrix The upper vertex is (0, b). The range of eccentricity is required!
• 7. 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
• 8. 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
• 9. Hyperbola has the same focus as ellipse x square + 2Y square = 20, its asymptote is 3x-y = 0, then hyperbolic equation is?
• 10. 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
• 11. Given that the square of hyperbola 2x - the square of 3Y = 18, what is the absolute value of the distance difference between a point and two focal points on the hyperbola, and what is the focal length? Answer the specific steps
• 12. Given that the absolute value of the distance difference between a point on the hyperbola and two focal points (- 2,0) and (2,0) is 2, then the hypohyperbolic equation is A 3 / 3 x ^ 2-y ^ 2 = 1 B x ^ 2-2 / 3 y ^ 2 C 3 / x ^ 2-y ^ 2 = - 1 D x ^ 2-3 of Y ^ 2 = - 1
• 13. Let ellipse C and hyperbola d have the same focus F1 (- 4,0), F2 (4,0), and the length of the major axis of ellipse is twice the length of the real axis of hyperbola. Try to find the trajectory equation of the intersection of ellipse C and hyperbola D
• 14. Ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) and hyperbola x ^ 2 / M-Y ^ 2 / N = 1 (m, n > 0) have common focus, F1, F2, P are their common points Ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) and hyperbola x ^ 2 / M-Y ^ 2 / N = 1 (m, n > 0) have common focus F1, F2 and P are their common points (1) Using B and N to express cos ∠ f1pf2 (2) Let s △ f1pf2 = f (B, n)
• 15. It is known that the ellipse x ^ 2 / M-Y ^ 2 = 1 (M > 1) and hyperbola x ^ 2 / n-y ^ 2 = 1 (n > 0) with the same two foci F 1 and F 2, P is one of their foci, then It is known that the ellipse x ^ 2 / M-Y ^ 2 = 1 (M > 1) and hyperbola x ^ 2 / n-y ^ 2 = 1 (n > 0) with the same two foci F 1 and F 2, and P is one of their foci, then the shape of triangle pf1f 2 is
• 16. If the hyperbola and ellipse 3x ^ 2 + 4Y ^ 2 = 48 are in common focus, and the real axis length is equal to 2, then the hyperbolic equation is?
• 17. To find the hyperbolic equation with the same focus as the ellipse X & # 178 / 16 + Y & # 178 / 8 = 1, the asymptotic equation is x ± √ 3Y = 0
• 18. It is known that the focus of the ellipse is on the x-axis, passing (2, root sign 3), and the eccentricity is root sign 3 / 2
• 19. The equation of the circle passing through the four intersections of ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 and Y ^ 2 / A ^ 2 + x ^ 2 / b ^ 2 = 1 (a > b > 0)
• 20. It is known that the parabola y = x * X-2 has four intersections with the ellipse y * y / 4 + X * x = 1. The equation of the circle with these four intersections is solved