To solve the hyperbolic equation with the vertex and focus of the ellipse as the focus and vertex respectively Urgent need process, thank you

Ellipse: a = 5, C = 3, then hyperbola C = 5, a = 3, that is, B = 4, so hyperbolic equation: x ^ 2 / 9 - y ^ 2 / 16 = 1

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1. Proposition p: "the equation x2 + y2m = 1 is an ellipse with focus on the y-axis", proposition q: "the function f (x) = 43x3-2mx2 + (4m-3) x-m increases monotonically on (- ∞, + ∞)". If P ∧ Q & nbsp; is a false proposition and P ∨ Q is a true proposition, the range of M is obtained
2. Given that the hyperbola and the square of ellipse X / 9 + the square of ellipse Y / 25 = 1 are in common focus, the sum of their eccentricities is 7 / 5, the hyperbolic equation is solved I'm sorry, but the difference between their eccentricities is 7 / 5
3. It is known that there are two intersections between a straight line passing through the right focus of hyperbola x2a2-y2b2 = 1 (a > 0, b > 0) and the right branch of hyperbola with an inclination angle of 45 degrees, then the value range of eccentricity e of hyperbola is______ ．
4. An example of hyperbola and its standard equation Examples are: When a shell explodes somewhere, the time of the explosion sound heard at a is 2 s later than that at B. It is known that the distance between AB and ab is 800 m and the sound velocity is 340 m / s. The equation of the explosion point curve is obtained (let the explosion point be p (x, y); establish a rectangular coordinate system so that AB is on the X axis, and point O coincides with the midpoint of line AB) In the process of solving the problem, there is one step "Because pa - Pb = 340 * 2 = 680 > 0, x > 0" I can't figure out why "pa - Pb = 340 * 2 = 680 > 0" can lead to "x > 0", and why to determine whether x is greater than 0?
5. Let the equation x2 / M-1 + Y2 / M + 3 = 1 denote hyperbola and find the value range of M one m-1>0,m+3>0, The solution is m > 1, M > - 3 The range of M is m > - 3 two (m-1)(m+3) Note that hyperbola is not ellipse. Explain the reasons and reasons for your choice
6. The equation x2-m of 3-m + Y2 of 2 = 1 represents hyperbola, then the value range of M
7. Given hyperbola X & # 178; - Y & # 178; = 4, straight line L: y = K (x-1), try to determine the value range of real number k, so that (1) The line L and hyperbola have two common points (2) Line L and hyperbola have and only have one common point (3) There is no common point between line L and hyperbola
8. It is known that a focal point of an ellipse is f (1,1), the corresponding collimator is x + y-4 = 0, and the eccentricity is √ 2 / 2
9. If the eccentricity of the ellipse is 12, a focal point is f (3,0) and the corresponding guide line is X-1 = 0, then the elliptic equation is______ ．
10. Given that the Quasilinear of the ellipse is x = 4, the corresponding focal coordinate is (2,0), and the eccentricity is 1 / 2, then the equation of the ellipse is?
11. : hyperbolic center in the origin, the focus on the X axis, through the point (2, - 3) and asymptote is y = positive and negative two-thirds x, autumn hyperbolic equation
12. The focus is (0,6) (0, - 6), and the hyperbolic standard equation is obtained through the point (2, - 5)
13. The standard equation of hyperbola with focus (0,6), (0, - 6) passing through points (2, - 5) Best to write the process! Thank you
14. The focus is (0, - 6), (0,6), and the hyperbolic standard equation passing through points (2, - 5) is () A. y220-x216=1B. x220−y216＝1C. y216−x236＝1D. x216−y236＝1
15. It has a common focus with the hyperbola x ^ 2 / 16-y ^ 2 / 9 = 1, and the hyperbolic equation can be solved through the point (- 3 radical 2,4)
16. The focus is on the x-axis, a is equal to twice the change sign 5, passing through the point a (- 5,2)
17. Find the standard equation of hyperbola (1) with focus on X-axis a = 2 √ 5 passing through point a (5,2)
18. Hyperbolic equation with a = 4, B = 3 and focus on X-axis I want to ask... X ^ 2 / 9-y ^ 2 / 16 = 1... Is this one true?
19. The standard equation of hyperbola with a = 4, B = 3 and focus on X axis is
20. If the distance from a point to the center of an equiaxed hyperbola is D, then the product of the distances from point P to two focal points is? I know the answer is d squared, online, etc