Know that elliptic equation, hyperbola and it have common focus, eccentricity reciprocal, find hyperbola formula? X ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1. If you use a / B to express the equation of open hyperbola, you don't need to calculate it later, and you don't want to deduce it yourself?

Know that elliptic equation, hyperbola and it have common focus, eccentricity reciprocal, find hyperbola formula? X ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1. If you use a / B to express the equation of open hyperbola, you don't need to calculate it later, and you don't want to deduce it yourself?

Very simple, several properties of elliptic hyperbola can be solved. The square of ellipse a minus the square of B equals the square of C. the hyperbola is a plus sign, and the eccentricity C is greater than a
Hyperbola C1 and hyperbola x2 / 2-y2 / 4 = 1 have the same asymptote, and pass through the point a (2, - √ 6), ellipse C2 takes the focus of hyperbola C1 as the focus, and the ellipse C2 has the same asymptote
The shortest distance between the point and the focus is √ 3, and the equations of hyperbola C1 and ellipse C2 are solved
Because the hyperbola C1 and x ^ 2 / 2-y ^ 2 / 4 = 1 have the same asymptote, we can set the equation of C1 as x ^ 2 / 2-y ^ 2 / 4 = K,
By substituting a coordinate, we can get 4 / 2-6 / 4 = k, and the solution is k = 1 / 2,
The equation of ^ 2 / y is reduced to C1 / 2 / 2-2 / y = 1 / 2
The focus of C1 is (- √ 3,0), (√ 3,0), so in the ellipse C2, C = √ 3,
The shortest distance between the point on the ellipse and the focus is a-c = √ 3,
So the solution is a = 2 √ 3, C = √ 3, so a ^ 2 = 12, B ^ 2 = a ^ 2-C ^ 2 = 9,
So the equation of ellipse C2 is x ^ 2 / 12 + y ^ 2 / 9 = 1
If the image of the linear function y = MX + (4M-2) of X passes through the first three or four quadrants, then the value range of M is that the value of Y increases and the value of M decreases
Because the image of the linear function y = MX + (4M-2) of X passes through the first three or four quadrants
So m > 0
4m-2
As for the image of the linear function y = MX + (4M-2) of X passing through the first three or four quadrants, the slope of the straight line k > 0, M > 0
The intercept of a straight line on the x-axis, (2-4m) / M > = 0, that is, 2-4m > = 0, M=
When the line y = KX + 1 intersects with the hyperbola 3x square minus y square = 1 and ab points, OA is perpendicular to ob when k is the value
You can just talk about ideas
But let me understand
Let a (x1, Y1) B (X2, Y2) OA be perpendicular to obx1x2 + y1y2 = 0x1x2 + (kx1 + 1) (kx2 + 1) = 0 (1 + K ^ 2) x1x2 + K (x1 + x2) + 1 = 0. Then we find the equation system of Weida's theorem y = KX + 13X ^ 2-y ^ 2 = 1, eliminate y, get the quadratic function of X, find X1 + x2x1x2 and substitute it into the above formula
Find the symmetric point of the image of the function f (x) = [cos (x) + sin (x)] sin (x)
f(x)=[cos(x)+sin(x)]sin(x)=cos(x)sin(x)+sin^2(x)=1/2sin2x+1/2(1-cos2x)
=√2/2[cos(p/4)sin2x-sin(p/4)cos2x]+1/2=√2/2sin(2x-p/4)+1/2
2x-p/4=kp
x=(4k+1)p/8
Symmetry point of image: ((4K + 1) P / 8,0) k is an integer
In order to make the image of function y = (2m-6) x + 2 pass through the first, second and third quadrants, the value range of M is
The general formula of a function is y = KX + B. when k is greater than 0. B is less than 0, it passes through one three four quadrants. So 2m-6 is greater than 0. M is greater than 3
2m-6
The left branch of the line y = KX + 1 and the square of hyperbola 3x - y = 1 intersects at point a and the right branch intersects at point B
(1) : find the value range of real number k?
(2) If the vector of OA is multiplied by the vector of OB = 0, the value of K is obtained?
(3) If a circle with the diameter of line AB passes through the coordinate origin, the equation of the circle is solved?
(1) The range of K is as follows
The critical point is the tangent of hyperbola (lower end) and straight line, that is, there is only one intersection point: KX + 1 = 3x2-1
Note: 3x2 is the square of 3x
How to count me? I haven't contacted for a long time. Anyway, I figure out the value of K
(2) The OA ob vector is 90 degrees
(3) The middle point C of line AB is the center of the circle, and the circle passes through the origin, that is, OC is perpendicular to ab
I'll figure it out by myself. I really forgot the details
Let f (x) = sin Wu x (x = 0), G (x) = cos Wu x (x = 1 / 2)
Find the value of G (1 / 4) + F (1 / 3) + G (5 / 6) + F (3 / 4)
g(1/4)+f(1/3)+g(5/6)+f(3/4)
=Cos (Wu / 4) + F (- 2 / 3) + 1 + G (- 1 / 6) + 1 + F (- 1 / 4) + 1
=√2/2-√3/2+1+√3/2+1-√2/2+1
=3.
thank you!
If the image of the function y = (2m-1) x and y = 3-m / X intersect one or three quadrants, then the value range of M is the same
These two functions don't seem to intersect one or three quadrants
Because if you want to cross the first and third quadrants, you must meet the following requirements at the same time:
2m-1>0 m
The straight line L: y = KX + 1 and hyperbola C: 3x ^ 2-y ^ 2 = 1 intersect at different points a and B
The left branch of the line L: y = KX + 1 and the hyperbola C: 3x ^ 2-y ^ 2 = 1 intersects a and the right branch intersects B
(1) Finding the value range of real number k
(2) Take the diameter of the ellipse as the origin
If y = KX + 1 and 3x & sup2; - Y & sup2; = 1 are eliminated simultaneously, we can get: (K & sup2; - 3) x & sup2; + 2kx + 2 = 0. According to Weida's theorem, XA + XB = 2K / (3-K & sup2;), xaxb = 2 / (K & sup2; - 3) because the left branch of the straight line L: y = KX + 1 and the hyperbola C: 3x ^ 2-y ^ 2 = 1 intersects a and the right branch intersects B  xaxb = 2 / (K & sup2;)
Substituting y = KX + 1 into 3x ^ 2-y ^ 2 = 1
We can get (3-K ^ 2) x ^ 2-2kx = 2
When k ^ 2 = 3, the equation has only one solution instead of two intersections of a and B. So it doesn't work
When k ^ 2 is not equal to 3, we can get: (3-K ^ 2) [x-k / (3-K ^ 2)] ^ 2 = 2 + K ^ 2 / (3-K ^ 2)
So [2 + K ^ 2 / (3-K ^ 2)] / (3-K ^ 2) is greater than 0
When 3-K ^ 2 is greater than 0, because K ^ 2 is greater than 0, expand [2 + K ^ 2 /
Substituting y = KX + 1 into 3x ^ 2-y ^ 2 = 1
We can get (3-K ^ 2) x ^ 2-2kx = 2
When k ^ 2 = 3, the equation has only one solution instead of two intersections of a and B. So it doesn't work
When k ^ 2 is not equal to 3, we can get: (3-K ^ 2) [x-k / (3-K ^ 2)] ^ 2 = 2 + K ^ 2 / (3-K ^ 2)
So [2 + K ^ 2 / (3-K ^ 2)] / (3-K ^ 2) is greater than 0
When 3-K ^ 2 is greater than 0, because K ^ 2 is greater than 0, [2 + K ^ 2 / (3-K ^ 2)] is greater than or equal to 2
So listing was established
When 3-K ^ 2 is less than 0, [2 + K ^ 2 / (3-K ^ 2)] is less than 0
That is: 6-k ^ 2 is greater than or equal to 0 can get the range of K
2. If a circle with diameter AB passes through the origin o, the lengths of AO and Bo are equal
Drawing can see that the hyperbola is symmetrical about the origin and about the Y axis
The straight line must pass through (0,1)
So when it is true that the line is parallel to 0, that is to say, the line passes through the axis of X
When a straight line passes through 0, it obviously has no focus with the hyperbola
So k = 0
That is, the straight line is y = 1, and the coordinates of points a and B can be obtained by substituting it into the equation