It is known that the equation of ellipse C1 is x24 + y2 = 1, the left and right focus of hyperbola C2 are the left and right vertex of C1 respectively, and the left and right vertex of C2 are the left and right focus of C1 respectively

It is known that the equation of ellipse C1 is x24 + y2 = 1, the left and right focus of hyperbola C2 are the left and right vertex of C1 respectively, and the left and right vertex of C2 are the left and right focus of C1 respectively

Let the equation of hyperbola C2 be x2a2-y2b2 = 1 (a > 0, b > 0), then A2 = 4-1 = 3, C2 = 4. From A2 + B2 = C2, B2 = 1 is obtained. Therefore, the equation of C2 is x23-y2 = 1
It is known that the elliptic equation is x ^ 2 / 9 + y ^ 2 = 1, make a straight line with an inclination angle of 30 degrees through the left focus, intersect the ellipse at two points a and B, and find the length of the chord ab
C ^ 2 = a ^ 2-B ^ 2 = 9-1 = 8, C = 2 √ 2, left focus coordinate f (- 2 √ 2,0), linear slope k = tan30 ° = √ 3 / 3, linear equation: y = √ 3 / 3 (x + 2 √ 2), substituting into elliptic equation, x ^ 2 / 9 + [√ 3 / 3 (x + 2 √ 2)] ^ 2 = 1,4x ^ 2 + 12 √ 2x + 15 = 0, according to Weida's theorem, X1 + x2 = - 3 √ 2, X1 * x2 = 15 / 4, according to the chord length
If the left focus angle of the ellipse x + 2Y = 4 is 60 degrees, then the length of the chord AB is
I'm in a hurry. I'm in a hurry. Thank you. I have to hand in my homework
According to the meaning: C ^ 2 = 2, the left focus of the ellipse is (- √ 2,0)
The slope of the line where the string AB is located is k = Tan θ = Tan 60 ° = √ 3
The linear equation of string AB: y = √ 3 x + √ 6
The linear equation where the string AB is located is combined with the elliptic equation to obtain the coordinates of two points AB, and the solution is: x = (- 6 √ 2 ± 4) / 7
Then: y = (- 6 √ 6 ± 4 √ 3) / 7 + √ 6
Length of string AB = 16 / 7
If | ab | ≤ 2, the value range of inclination angle α of straight line AB can be obtained
Let a: x1, y1b: X2, Y2 (y2-y1) ^ 2 + (x2-x1) ^ 2 ≤ 4 ∵ Y1 = K (x2-x1) ^ 2 = K (x1 + √ 2) by | ab | ≤ 2, AB ^ 2 ≤ 4, let a: x1, y1b: X2, Y2 (y2-y1) ^ 2 + (x2-x1) ^ 2 ≤ 4 ∵ Y1 = K (x1 + √ 2) y2 = K (x2 + √ 2)}
If sin (π 6 − a) = 13, then the value of COS (2 π 3 + 2a) is ()
A. -79B. -13C. 13D. 79
Cos (2 π 3 + 2 α) = - cos (π 3-2 α) = - cos [2 (π 6 − α)] = - [1-2sin2 (π 6 − α)] = - (1-29) = - 79, so a is selected
If y = (2,2) is a cubic function, then M-1 is a cubic function?
Because y = (m-2) X-1 passes through the second, third and fourth quadrants, so m-2 ＜ 0.. so m ＜ 2
The straight line y = KX + 1 and hyperbola 3x ^ 2 - y ^ 2 = 1 intersect at A.B
When k is a real number, points a and B are on the left branch of hyperbola
Two equations are established
y=kx+1
3x²-y²=1
The elimination of y was as follows
(3-k²）x²-2kx-2=0
According to the meaning of the title:
Δ>0
2k/(3-k²）0
Solution: √ 3
If sin (A-wu / 6) = 1 / 3, then the value of COS (2a + Wu / 6) is equal to
Need process
Let x = a - π / 6
a=x+π/6
sinx=1/3
sin²x+cos²x=1
So cosx = ± 2 √ 2 / 3
So the original formula = cos (2x + π / 3 + π / 6)
=cos(2x+π/2)
=-sin2x
=-2sinxcosx
=±4√2/9
The image of a function y = 2x-m + 1 passes through 1, 2, 3 quadrants, and the value range of M is obtained
-M+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. (1) when k = 2, find the length of ab;
(2) Is there a real number k such that the circle with the diameter of line AB passes through the coordinate origin?
(1) K = 2, then y = KX + 1 = 2x + 1, the line L: y = 2x + 1 and hyperbola C: 3x ^ 2-y ^ 2 = 1 intersect at two different points a and B, so let a (x1,2x1 + 1) and B (x2,2x2 + 1)
By substituting y = 2x + 1 into 3x ^ 2-y ^ 2 = 1, we can get x ^ 2 + 4x + 2 = 0, and then we can get X1 + x2 = - 4, (x1) (x2) = 2
|Ab | = radical {(x2-x1) ^ 2 + [(2x2 + 1) - (2x1 + 1)] ^ 2} = radical [5 * (x2-x1) ^ 2]
=Root {5 * [(x2 + x1) ^ 2-4 (x1) (x2)]} = 2 times root 10;
(2) A (x1, kx1 + 1) and B (X2, kx2 + 1) can be obtained from (1)
Substituting y = KX + 1 into 3x ^ 2-y ^ 2 = 1, we get: (3-K ^ 2) x ^ 2-2kx-2 = 0
X1 + x2 = (2k) / (3-K ^ 2), (x1) (x2) = 2 / (k ^ 2-3), discriminant = (- 2K) ^ 2 + 8 (3-K ^ 2) = 24-k ^ 2 > 0, then
|Ab | = radical {(x2-x1) ^ 2 + [(kx2 + 1) - (kx1 + 1)] ^ 2} = radical [(1 + K ^ 2) * (x2-x1) ^ 2]
=Root {(1 + K ^ 2) * [(x2 + x1) ^ 2-4 (x1) (x2)]} = 2 times root {[(6-k ^ 2) (1 + K ^ 2)] / [(3-K ^ 2) ^ 2]};
Therefore, the radius of the circle [^ 2] / (K-3)] (K-2) / (K-2)] (K-3) is the root of the equation
If the coordinate origin is on this circle, then [K / (3-K ^ 2)] ^ 2 + [3 / (3-K ^ 2)] ^ 2 = [(6-k ^ 2) (1 + K ^ 2)] / [(3-K ^ 2) ^ 2]
That is, K ^ 2 = 3 or K ^ 2 = 1
In conclusion, there is a real number k such that the circle with the diameter of line AB passes through the coordinate origin
1. When k = 2, y = 2x + 1, 3x ^ 2-y ^ 2 = 1
A. The coordinates of two points B are respectively: (- 2 + √ 6, - 3 + 2 √ 6), (- 2 - √ 6, - 3-2 √ 6)
Length of AB = √ [(- 2 + √ 6 + 2 + √ 6) ^ 2 + (- 3 + 2 √ 6 + 3 + 2 √ 6) ^ 2] = 2 √ 30
(2) Whether there is a real number k, so that the circle with the diameter of line AB passes through the coordinate origin, that is, a and B are symmetrical about the origin
X1 + x2 = 0, Y1 + y2 =... Expansion
1. When k = 2, y = 2x + 1, 3x ^ 2-y ^ 2 = 1
A. The coordinates of two points B are respectively: (- 2 + √ 6, - 3 + 2 √ 6), (- 2 - √ 6, - 3-2 √ 6)
Length of AB = √ [(- 2 + √ 6 + 2 + √ 6) ^ 2 + (- 3 + 2 √ 6 + 3 + 2 √ 6) ^ 2] = 2 √ 30
(2) Whether there is a real number k, so that the circle with the diameter of line AB passes through the coordinate origin, that is, a and B are symmetrical about the origin
x1+x2=0,y1+y2=0
By substituting y = KX + 1 into the curve equation, we get the following results:
(K-3) x ^ 2-2kx-2 = 0 should have X1 + x2 = 0, that is: 2K / (K-3) = 0 to get k = 0
When k = 0, y = 1, does not conform to Y1 + y2 = 0
To sum up, there is no real number k, so that the circle with the diameter of line AB passes through the coordinate origin. Ask: when k = 2, there are y = 2x + 1, 3x ^ 2-y ^ 2 = 1, do you get the answer? The coordinates of a and B are respectively: (- 2 + √ 2, - 3 + 2 √ 2), (- 2 - √ 2, - 3-2 √ 2). Please help me to calculate again, thank you!