![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
Given the hyperbola 2x2-y2=2, the straight line L passing through the point P (2,1) intersects the hyperbola at two points A and B. If the straight line AB is parallel to the y-axis, the length of the segment AB is obtained. Given the hyperbola 2x2-y2=2, the straight line L passing through the point P (2,1) intersects the hyperbola at the two points A and B. If the straight line AB is parallel to the y-axis, the length of the segment AB is obtained.
From the meaning of the title, replace the straight line x=2 into hyperbola 2x2-y2=2, and y=±
6,
Length of line segment AB is 2
6.
According to the meaning, replace the straight line x=2 into hyperbola 2x2-y2=2, then y=±
6,
Length of line segment AB is 2
6.
Given the hyperbolic points A (-2,4), B (4,4), one focal point which is F1(1,0), find the trajectory equation of the other focus F2. Given the hyperbolic points A (-2,4) and B (4,4), one focus of which is F1(1,0), find the trajectory equation of the other focus F2.
Hyperbolic points A (-2,4) and B (4,4), one of whose focal points is F1(1,0),
AF1|=|BF1|=5,
According to the definition of hyperbola,||AF1|-|AF2||=||BF1|-|BF2||, i.e.|5-|AF2||=|5-|BF2||,
(1) When 5-|AF2|=5-|BF2|, i.e.|AF2|=|BF2|,
The locus of the focus F2 is the mid-perpendicular of the line segment AB, whose equation is x =1(y =0 or 8),
(2) When 5-|AF2|=|BF2|-5, i.e.|AF2 BF2|=10>6,
The locus of the focus F2 is an ellipse with A and B as the focus and the major axis of the ellipse is 10.
Its center is (1,4), a=5, c=3, b2=25-9=16,
Its equation is (x-1)2
25+(Y-4)2
16=1(Y =0)
To sum up, the trajectory equation of the other focus F2 is: x=1(y=0 or 8) or (x-1)2
25+(Y-4)2
16=1(Y =0).
Let P be hyperbolic x 2 4-Y2=1, O is the origin of coordinates, M is the midpoint of line segment OP, then the trajectory equation of point M is ______. Let P be hyperbolic x 2 4-Y2=1, O is the coordinate origin, M is the midpoint of line segment OP, then the trajectory equation of point M is ______.
Let M (x, y), then P (2x,2y) be substituted into the hyperbolic equation and x2-4y2=1.
Trajectory equation x2-4y2=1.
Answer: x2-4y2=1
Let M (x, y), then P (2x,2y) be substituted into the hyperbolic equation to obtain x2-4y2=1.
Trajectory equation x2-4y2=1.
Answer: x2-4y2=1
Given that the point P of A (0,1) B (0,-1) C (1,0) satisfies the vector AP* vector BP=2 vector PC^2,(1) Find the trajectory equation of P
If the coordinate of point P is (x, y), then because vector AP*vector BP=2vector PC^2,
(X, y-1)*(x, y+1)=2*(x-1, y)^2
Available:
X^2-4x+4+y^2=1;
So (x-2)^2+y^2=1
An equation that knows to be a circle;
Given points A (-2,0), B (2,0), the moving point P on curve C satisfies vector AP multiplied by vector BP =-3 (1) Equation of curve C. (2) If the straight line L passing through the fixed point M (0,-2) has an intersection with the curve c, find the value range of the slope of the straight line L. (3) If the moving point Q (x, y) is on the curve c, find the value range of U=y+2/x. Given points A (-2,0), B (2,0), the moving point P on curve C satisfies vector AP multiplied by vector BP =-3 (1) The equation of curve C. (2) If the straight line L passing through the fixed point M (0,-2) has an intersection with the curve c, find the value range of the slope of the straight line L. (3) If the moving point Q (x, y) is on the curve c, find the value range of U=y+2/x.
1> Given point A (-2,0), B (2,0) Let P be point =(x, y) So, vector AP =(x+2, y), vector BP =(x-2, y) So, quantity AP is multiplied by vector BP = x2+y2-4=-3, i.e. x2+y2=1 So, equation of curve C: x2+y2=1 Given point M =(0,-2) So, set straight line l: y=kx-2, and...
1> Given point A (-2,0), B (2,0) Let P be point =(x, y) So, vector AP =(x+2, y), vector BP =(x-2, y) So, quantity AP is multiplied by vector BP = x2+y2-4=-3, i.e. x2+y2=1 So, equation of curve C: x2+y2=1 Given fixed point M =(0,-2) So, set straight line l: y=kx-2, and...
1> Given point A (-2,0), B (2,0) Let P be point =(x, y) So, vector AP =(x+2, y), vector BP =(x-2, y) So, quantity AP is multiplied by vector BP = x2+y2-4=-3, i.e. x2+y2=1 So, equation of curve C: x2+y2=1 Given point M =(0,-2) set straight line l: y=kx-2, and...
Given that the area of triangle ABC is 3 and satisfies 0≤vector AB·vector AC≤6, let the included angle of vectors AB and AC be θ 1. Calculate the value range of θ 2. Find the maximum and minimum values of the function (θ)=2sin^2(π/4)-√3 cos2θ
1. Because area of triangle ABC=(ABXAC) sinθ/2=3
ABXAC sinθ=6--> sinθ=6/ABXAC.(1)
And 0≤vector AB·vector AC≤6
0≤ABxACcos 6--->0≤cos 6/ABxAC.(2)
(1) Substitute into (2)
0≤Cos sinθ
So π/4/2
2.
Because π/4/2
Cos2 0
When θ=π/4
ƒ(θ)=2Sin^2(π/4)-√3 cos2θ=2
Is minimum
When θ=π/2
ƒ(θ)=2Sin^2(π/4)-√3 cos2θ=1 3
Is the maximum
RELATED INFORMATIONS
- 1. The angles of the three vectors of abc are known to be equal in two, and the module of vector a is equal to the module of vector b is equal to 1, the module of vector c is equal to 3, and the module of vector a+b+c is found. It is known that the angles of the three vectors of abc are equal in two, and the module of vector a is equal to the module of vector b is equal to 1, and the module of vector c is equal to 3, so as to find the module of vector a+b+c.
- 2. In triangular ABC, the vector AB=a, the vector AC=b, ab <0, S triangular ABC=15/4,|a|=3, |B |=5, then angle BAC = It is known that in triangle ABC, vector AB=a, vector AC=b, ab <0, S triangle ABC=15/4,|a|=3, |B |=5, then angle BAC =
- 3. If the angle between vector a and vector b is 120° and a =4'|b|=2', then a·b=?
- 4. Linear algebra: If the matrices a and b are equal, then the row vector group of a is equivalent to the row vector group of b Why? Linear algebra: If matrices a and b are equal, then the row vector group of a is equivalent to the row vector group of b. Why?
- 5. If the vector a =(k,1), b =(4, k) is collinear and the directions are opposite, then k =() There is something wrong with the answer and option, so I won't write it. But I want to ask you if it is 2 or -2, and the answer makes me confused!
- 6. If the function f (x)=x2+ax is even, then a =
- 7. Known vector A =(3,1), B =(1,3), C=(k,2), if ( A− C)⊥ B, k=______. Known vector A =(3,1), B =(1,3), C=(k,2), if ( A− C)⊥ B then k =______.
- 8. Given that the angle between vector a and vector b is 120 degrees and the module of vector a = the module of vector b =4, then the value of vector b multiplied (2 multiplied by vector a + vector b) is:
- 9. Vector a=(cosx, sinx), vector (b=-√3,-1), then what is the maximum and minimum values of 2a-b?
- 10. Given vector a=(3,0), b=(K,5), and the angle between vector a and vector b is 3π/4, then the value of k is
- 11. In the Cartesian coordinate plane xOy, given the point A (3,2), the point B moves on the circle x^2+y^2=1, the moving point P satisfies the vector AP=vector PB, then the trajectory of the point P is? ... In the Cartesian coordinate plane xOy, given the point A (3,2), the point B moves on the circle x^2+y^2=1, the moving point P satisfies the vector AP=vector PB, then the trajectory of the point P is? Circular elliptic parabolic straight line
- 12. Point A (-1,0) B (1,4) satisfies vector PA·vector PB=4, and the trajectory equation of point P is obtained. If the point Q is a symmetric point of P with respect to the straight line y=2(x-4), find the trajectory equation of the moving point Q Point A (-1,0) B (1,4) satisfies the vector PA·vector PB=4, and the trajectory equation of point P is obtained. If the point Q is a symmetric point of P with respect to the straight line y=2(x-4), find the trajectory equation of the moving point Q
- 13. Given vector OA=(3,-4), vector OB=(6,-3), vector OC=(5-x,-3-y), if A, B, C can form a triangle, find the condition that x, y satisfies.
- 14. Given vector a=(1, x), vector b=(x^2+x,-x), m is constant and m≤-2 Given vector a=(1, x), vector b=(x^2+x,-x), m is constant and m≤-2 Let a · b +2> m ((2/(a · b))+1). How did x get? Given vector a=(1, x), vector b=(x^2+x,-x), m is constant and m≤-2 Given vector a=(1, x), vector b=(x^2+x,-x), m is constant and m≤-2 Let a · b +2> m ((2/(a · b))+1). How was x found? Given vector a=(1, x), vector b=(x^2+x,-x), m is constant and m≤-2 Given vector a=(1, x), vector b=(x^2+x,-x), m is constant and m≤-2 Let a·b+2> m ((2/(a·b))+1). How did you get x?
- 15. Given that point D is the midpoint of the edge BC of the triangle, there is a point P in the plane where the triangle ABC is located, which satisfies vector PA + vector BP + vector CP =0, The module of vector PA is divided by the module of vector PD is equal to λ, and the value of λ is obtained.
- 16. Given that M={ a|a=(2 1,-2-2), R}, N={ b|b=(3 2,6 1), R}, is a set of vectors, then M∩N= Options have four A {c|c (-2,1)} B {c|c (2,1)} C {c|c (-2,1)} D {c|c (2,1)}
- 17. Vector a+b=(-2,-1), a-b=(4,-3), the angle between vector a, b is
- 18. What is the difference between quantity product and vector product? Why the quantity product is perpendicular if the vector a* vector b=0 and the vector a* vector b=0 if the vector product is parallel What is the area of the quantity product and the vector product? Why is the quantity product vertical if the vector a* vector b=0 and the vector product parallel if the vector a* vector b=0?
- 19. The vector a=(cosα, sinα), b=(cosβ, sinβ),0<β<α If |a-b|=root2 find a perpendicular b let c=(0,1) if a+b=c find the value of The vector a=(cosα, sinα), b=(cosβ, sinβ),0<β<α If |a-b|=rootoot 2, find a perpendicular b Let c=(0,1) If a+b=c, find the value of
- 20. Given vector a=(x^2-3,1) b=(x,-y), when x <2 in absolute value, there is a⊥b, when x≥2 in absolute value, a‖b Given vector a=(x^2-3,1) b=(x,-y),(where y and x are not equal to 0), when x <2 in absolute value, there is a⊥b, when x≥2 in absolute value, a‖b, find function y=f (x)