Given y = (m ^ 2-m-2) x ^ m ^ 2-2m-1, when m is, y is the inverse proportion function of X

Given y = (m ^ 2-m-2) x ^ m ^ 2-2m-1, when m is, y is the inverse proportion function of X

Y is the inverse proportional function of X, then the exponent of X is equal to - 1, and the coefficient is not equal to 0. From m ^ 2-2m-1 = - 1, the solution is m = 2 or M = 0. Because m ^ 2-m-2 should not be equal to 0, M = 2 is rounded off, so m = 0 is the required value
Given the X-type elliptic standard equation, how to set the hyperbolic equation with its common focus? Why?
Given y = (2m-1) x ^ m ^ 2-3, when m is what value, it is an inverse proportional function
The condition is: 2m-1 is not 0, (coefficient is not 0)
M ^ 2-3 = - 1 (index is - 1)
m^2=2
M equals the positive and negative root sign 2,
Given that the two focal coordinates of the ellipse are (0, - 4) (0,4) and a = B, the standard equation is solved
Child, as a freshman, I want to say, study hard, because the focus coordinate is 0,4, so C = 4, because a = B, so the square of a plus the square of B = the square of C, so a = b = 2, my mobile phone can't input equation, you can write the equation yourself, you can't ask me, as long as I can, I will help you
If y = x ^ 2m-1 is an inverse scale function, then the value of M is?
Because there is another expression of inverse proportion function, that is, y = x Λ - 1, which is evolved from y = 1 / X
∵y=x^2m-1
∴2m-1=-1
∴m=0
2m-1=-1,
M=0
It is known that the two focal coordinates of an ellipse are (0, - 2) and (0,2) respectively and pass through (3 / 2,5 / 2), and the standard equation of the ellipse is obtained
c = 2
2a = √[(0 - 3/2)^2 + (2 - 5/2)^2] + √[(0 - 3/2)^2 + (-2 - 5/2)^2]
= √(10/4]) + √ (90/4)
= 2√10
So a = √ 10
That is, B = √ (10 - 4)
= √6
So the equation of ellipse is x ^ 2 / 6 + y ^ 2 / 10 = 1
。。。 It's very simple
If point (3,4) is a point on the image of inverse scale function y = M & # 178; + 2m + 1 / x, then the function image must pass through point ()
Explain why
Yes (- 3, - 4)
Because y = M & # 178; + 2m + 1 / X
M & # 178; + 2m + 1 = (M + 1) ^ 2 positive
(4,3) points
Because the inverse proportion function is hyperbolic and does not intersect the X and Y axes, the two curves must be in the first and third quadrants or the second and fourth quadrants, (3,4) in the first quadrant, so the other straight line must pass through the third quadrant (- 3, - 4),
M & # 178; + 2m + 1 of the molecule is greater than or equal to 0. Can you see that
So the function is y = K / x, k > = 0. The image is in one or three quadrants
When x goes to the opposite number, y goes to the opposite number.
So it must go through (- 3, - 4)
If I'm right, choose me
A standard elliptic equation with F (0 - 3) as the focus
Such as the title
C=3
c/a=1/2 a=6
b^2=36-9=27
y^2/36+x^2/27=1
If the point (3,4) is a point on the inverse scale function y = M2 + 2m − 1x image, then the function image must pass through the point ()
A. （2，6）B. （2，-6）C. （4，-3）D. （3，-4）
∵ point (3,4) is a point on the inverse scale function y = M2 + 2m − 1X, and ∵ point (3,4) satisfies the inverse scale function y = M2 + 2m − 1X, ∵ 4 = M2 + 2m − 13, that is, M2 + 2m-1 = 12, ∵ point (3,4) is a point on the inverse scale function y = 12x, ∵ xy = 12; a, ∵ x = 2, y = 6, ∵ 2 × 6 = 12
It is known that the focus of ellipse C is on the x-axis, the point P (1.3 / 2) is on the ellipse, and the eccentricity is 1 / 2. The standard equation of ellipse C is obtained
A ^ 2 = 3B ^ 2
Let x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1. (2)
Substituting the coordinates of point P into (2) yields 1 / A ^ 2 + 9 / 4B ^ 2 = 1. (3)
(1) (3) get a ^ 2 and B ^ 2 simultaneously, and then get the elliptic standard equation. The rest can be solved by yourself, which is good for you