Eccentricity of elliptic hyperbola Literal expression of hyperbolic eccentricity

Eccentricity of elliptic hyperbola Literal expression of hyperbolic eccentricity

The distance from a point on the elliptic hyperbola to x = a ^ 2 / C is more centrifugal than the distance from the point to the focus C
Since you didn't explain what to ask, I can only add
If the ellipse x ^ 2 / m ^ 2 + y ^ 2 / N ^ 2 = 1 and the hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 have the same focus, and F1F2 P is one of their intersection points, then the value of Pf1 times PF2 is?
Utilization definition:
｜PF1｜+｜PF2｜＝2|m|
||PF1|-|PF2||=2a
By subtracting the square of the two formulas, we get the following result:
4|PF1||PF2|=4(m^2-a^2)
|PF1||PF2|=m^2-a^2
If the line y = KX is tangent to the curve y = 2ex, then the real number K=______ .
Let the tangent point be (x0, Y0), then Y0 = 2ex0, ∵ y ′ = (2ex) ′ = 2ex, ∵ tangent slope k = 2ex0, and the point (x0, Y0) is on the straight line, then Y0 = kX0, that is, 2ex0 = 2ex0, the solution is X0 = 1, ∵ k = 2E
The result of simplifying √ (2 + Cos2 sin squared 1) is
2+cos2-sin²1=1+cos2+1-sin²1=2cos²1+cos²1=3cos²1
Because one radian is about 57.6 degrees
So cos1 > 0
Then the original formula = √ 3cos1
If the image of the first-order function y = (m-3x + 2m-2) passes through one two four quadrants and M is an integer 1, the value of M can be obtained
The image of linear function y = (M-3) x + 2m-2 passes through one two four quadrants
∴﹛m-3＜0 m＜3
2m-2＞0 m＞1
∴1＜m＜3
∵ m is an integer
∴m=2
If the line y = KX + 1 intersects the circle x2 + Y2 + KX + my-4 = 0 at two points m and N, and m and N are symmetric with respect to the line x + 2Y = 0, then the real number K + M = ()
A. -1B. 1C. 0D. 2
From the meaning of the question, we can get that ∵ the line y = KX + 1 intersects with the circle x2 + Y2 + KX + my-4 = 0 at two points m and N, and m and N are symmetric with respect to the line x + 2Y = 0, ∵ the line x + 2Y = 0 is the vertical line of the line Mn, and we get k · (- 12) = - 1, and the solution is k = 2, so the circular equation is x2 + Y2 + 2x + my-4 = 0, and the coordinates of the center of the circle are (− 1, − m2). Substituting (− 1, − m2) into x + 2Y = 0, we get m = - 1, and we get K + M = 1
Root 2 + Cos2 sin square 1
All the numbers are in the root
cos2=1-2sin²1
So the formula under the root sign = 2 + 1-2sin & sup2; 1-sin & sup2; 1
=3(1-sin²1)
=3cos²1
Zero
cos1
Radical (2 + 2cos1 ~ 2-1 + cos1 ~ 2-1) = radical cos1 ~ 2
=cos1
If the image of a linear function y = (2-m) x + m passes through the first, second and fourth quadrants, what is the value range of M? If its image does not pass through the second quadrant, what is the value range of M?
If the solution passes through quadrant one two four, then there is 2-mo, and the solution is m > 2
If the function image does not pass through the second quadrant, then there are 2-m ≥ 0 and m ≤ 0, and the solution is m ≤ 0
Why do you write this
1. a=2-m b=m
Because it passes through quadrant 24, A0 and M > 0
Because m > 0, M > 2, M > 2
2. If a > 0 and B is less than or equal to 0 2-m > 0 m without passing through 2 quadrants
If K is a real number and K [- 2,2], what is the probability that two straight lines passing through point a (1,1) are tangent to the circle x ^ 2 + y ^ 2 + kx-2y - (5K / 4) = 0
1/4
First, a polynomial is a circle, so you get - 4
In fact, this problem, as long as you find out the value range of K when a point is outside the circle.
Because there must be two lines tangent to the circle passing through point a, so point a is outside the circle.
That is, 2 + K-2 - (5K / 4) > 0 (not equal to 0, only one line is tangent to 0)
K
Sin square x = 8 / 5, find Cos2 (π / 4-2x)
In fact, I want to ask how sin2x can be greater than 1