Some problems on the standard equation of hyperbola 1. The focal coordinates are (0, - 6), (0,6) and (2, - 5) respectively 2. Eccentricity e = 4 / 3, imaginary axis length is 2 times root 7, focus on Y axis 3. The focal length is 10, the distance between two vertices is 6, and the vertex is on the X axis

Some problems on the standard equation of hyperbola 1. The focal coordinates are (0, - 6), (0,6) and (2, - 5) respectively 2. Eccentricity e = 4 / 3, imaginary axis length is 2 times root 7, focus on Y axis 3. The focal length is 10, the distance between two vertices is 6, and the vertex is on the X axis

In the first question, we can get a & # 178; + B & # 178; = 36 from the meaning of the question, and then B & # 178; can be expressed by a & # 178; and can be solved by bringing in the coordinates
In question 2, B can be seen directly, and at the same time, it also tells you the centrifugal rate, so the AC relation can be known,
Question 3, the topic has told you the focal length and vertex distance, does not mean that a and C have told you Just calculate B directly

It is known that the proposition p: equation x ^ 2 / 2 + y ^ 2 / M = 1 represents an ellipse with focus on the Y axis; proposition q: F (x) = 4x ^ 3 / 3-2mx ^ 2 + (4m-3) x-m monotonically increases on (negative infinity, positive infinity), if (non-p) / Q is true, the value range of M is obtained

4m^2-16*（4m-3）＜0,m

The particle moves along a straight line, the initial velocity V0, the acceleration a = - K, the root sign V, K are normal numbers, and the time required for the particle to be completely stationary and the moving distance during this period are calculated

When the particle moves along a straight line, initial velocity V0, acceleration a = - K √ V, K is the normal number DV / dt = - K √ V DV / √ v = - KDT, the time required for √ v = √ V0 - kt / 2 particle to be completely stationary v = 0 t = (2 √ VO) / K √ v = √ VO - kt / 2 DX / 4T = v = VO - kt √ VO + K & sup2; T & sup2; / 4 x (T) = vot-2k √ VOT & sup2; + K & su

It is known that proposition p: equation x2m + y2 = 1 represents an ellipse with focus on Y axis; proposition q: there are two intersections between straight line y = X-1 and parabola y = MX2. (1) if proposition q is a true proposition, find the value range of real number m; (2) if there is only one true proposition between proposition p and Q, find the value range of real number M

Proposition p: the equation x2m + y2 = 1 denotes that the ellipse with focus on the y-axis ⇔ 0 ＜ m ＜ 1. (1) if proposition q: there are two intersections between straight line y = X-1 and parabolic y = MX2 ⇔ MX2 = X-1, there are two intersections ⇔ mx2-x + 1 = 0, there are two different real roots, m ≠ 01-4m ＞ 0, m ＜ 14 and m ≠ 0, so the value range of M is (- ∞, 0) ∪ (0, 14) (2) if there is and only one true proposition between proposition p and Q, then p True Q false or P false Q true, if P true Q false, then 14 ≤ m < 1; if P false Q true, then M < 0. So the value range of M is (- ∞, 0) ∪ [14,1]

It is proved that if a function takes any positive number as a period, then the function is a constant

To the contrary
Suppose there are two different points, f (x1) = a, f (x2) = B,
x1>x2, a≠b
Let t = x1-x2, then t is a positive number and it should be periodic
So f (x1 + T) = f (x1)
That is, f (x2) = f (x1)
b=a,
contradiction
So this function is constant

To solve the standard equation of hyperbola in mathematics of senior two: through two points a (negative 7.6 times root sign 2) and B (2 times root sign 7.3), to find the detailed explanation, especially the process, to simplify the part

If the focus cannot be determined, let the standard equation be: x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1,
(-7)^2/a^2-(-6√2)^2/b^2=1,
49/a^2-72/b^2=1.(1),
28/a^2-9/b^2=1.(2),
(2) Formula * 8 - (1)
175/a^2=7,
a^2=25,
a=5,
Substituting (2), 5 √ 3,
The hyperbolic equation is: x ^ 2 / 25-y ^ 2 / 75 = 1,
Let the hyperbolic equation be y ^ 2 / b ^ 2-x ^ 2 / A ^ 2 = 1,
72/b^2-49/a^2=1.(3),
9/b^2-28/a^2=1.(4),
(3) Formula - (4) formula * 8,
175/a^2=-7,
a^2=-25,
There is no real root, so the focus is not on the Y axis
The hyperbolic equation is as follows:
x^2/25-y^2/75=1.

It is proved that if the function f (z) is analytic in D and f '(z) = 0 in D, it is proved that f (z) must be a constant in D

Look at the picture:

Hyperbolic standard equation of mathematics in Senior Two
Hyperbolic standard equation passing through points (- 7, - 6, root 2) and (2, root 7, - 3)

Let the hyperbolic standard equation be MX ^ 2 + NY ^ 2 = 1 generation point (- 7, - 6 root sign 2) point (2 root sign 7, - 3) into the equation
49m+72n=1
28m + 9N = 1
21m+63m=0
M = - 3N
M = 1 / 25, n = - 1 / 75
Hyperbolic standard equation x ^ 2 / 25-y ^ 2 / 75 = 1

For a non-zero constant a function y = f (x) = - 1 / F (x), how to prove that there exists a period of 2A?

To think about it, you just need to prove that f (x + 2a) = f (x)
prove:
Since f (x) = - 1 / F (x), the,
F(X)*F(X)=-1,
F(X+A)*F(X+A)=-1
F(X)*F(X)=F(X+A)*F(X+A)
(F(X+A)+F(X))(F(X+A)-F(X))=0
F (x + a) = - f (x) or F (x + a) = f (x)
If f (x + a) = f (x), f (x) period is a, then 2a is its period
If f (x + a) = - f (x), f (x + 2a) = - f (x + a) = f (x), then 2a is its period
So f (x) has a period of 2A

Standard equation of hyperbola
Find the standard equation of hyperbola suitable for the following conditions:
The asymptote equation is y = + - 3 / 5x, and the focus coordinate is (+ - radical 2,0)
Especially the conversion, I can do, just don't know how to solve that answer, there are extra points

According to the meaning of the question, we know that hyperbola is the standard form, symmetric about X and Y axes, and the focus coordinate is on X axis, so let the hyperbola equation be: X & # 178; / A & # 178; - Y & # 178; / B & # 178; = 1, because the focus coordinate is (+ - radical 2,0), so a & # 178; + B & # 178; = 2, and the asymptotic equation be y = + - 3 / 5x, that is, let X & # 17