Given that the vertex BC of triangle ABC is on the ellipse = 1, and the edge BC passes through one focus of the ellipse, and vertex A is another focus of the ellipse, then the perimeter of △ ABC is

Given that the vertex BC of triangle ABC is on the ellipse = 1, and the edge BC passes through one focus of the ellipse, and vertex A is another focus of the ellipse, then the perimeter of △ ABC is

Ellipse (x ^ 2) / (a ^ 2) + (y ^ 2) / (b ^ 2) = 1 2a is the distance from any point to two focal points, then C △ = BF1 + BF2 + CF1 + CF2 = 4A elliptic equation, you can't find the amount
It is known that the center of the ellipse W is at the origin, the focus is on the x-axis, the length of the major axis is 4, the eccentricity is √ 6 / 3, and the vertices a, B and C of Δ ABC are on the ellipse,
C is on L: y = x + 2, and ab ‖ L
(1) Elliptic equation
(2) When AB passes through the coordinate origin, find the length of AB and the area of △ ABC
(3) When ∠ ABC = 90 ° and AC is maximum, the linear equation of AB is obtained
(1) Elliptic equation
Let x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1
The focus is on the x-axis, the length of the long axis is 4, and the eccentricity is √ 6 / 3
A=2
√(a^2-b^2)/a=√6/3
(2^2-b^2)/2^2=6/9
b^2=4/3
Elliptic equation: x ^ 2 / 4 + y ^ 2 / (4 / 3) = 1, or writing: x ^ 2 + 3Y ^ 2-4 = 0
(2) When AB passes through the coordinate origin, find the length of AB and the area of △ ABC
The vertices a, B, C of Δ ABC are on the ellipse, C is on L: y = x + 2, ab ‖ L,
The slope of the line where AB is located is KAB = ki = 1
When AB passes through the coordinate origin, the linear equation of AB is y = x, substituting x ^ 2 + 3Y ^ 2-4 = 0 to get x ^ 2 + 3x ^ 2-4 = 0,
x1=-1,x2=1
y1=-1,y2=1
AB=√((x2-x1)^2+(y2-y1)^2}=2√2
∵ I is parallel to AB, C is on I, AB passes through the origin,
The height of ABC = the distance from the origin to the line I, H = A / √ 2 = 2 / √ 2 = √ 2
| triangle ABC area = 1 / 2 | ab | * H = 1 / 2 * 2 √ 2 * √ 2 = 2
(3) When ∠ ABC = 90 ° and AC is maximum, the linear equation of AB is obtained
C substitutes y = x + 2 into x ^ 2 + 3Y ^ 2-4 = 0 in L: y = x + 2
x^2+3(x+2)^2-4=0
4x^2+12x+8=0
(x+2)(x+1)=0
x1=-2,x2=-1
y1=-2+2=0,y2=-1+2=1
Coordinates of point C (- 2,0), or (- 1,1)
∠ABC=90°,kBC=-1/kAB=-1/1 = -1
When the coordinate of point C (- 2,0), the linear equation of BC is y-0 = - 1 (x - (- 2)), that is, y = - X-2
Substituting y = - X-2 into x ^ 2 + 3Y ^ 2-4 = 0, x ^ 2 + 3 (- X-2) ^ 2-4 = 0, the solution is X1 = - 2, X2 = - 1
X = - 2 is the abscissa of point C
The abscissa x = - 1 and the ordinate y = - (- 1) - 2 = - 1 of point B, i.e. B (- 1, - 1)
The equation of AB is Y - (- 1) = 1 * (x - (- 1)), that is, y = X
Substitute y = x into x ^ 2 + 3Y ^ 2-4 = 0 to get 4x ^ 2 = 4, x = ± 1
X = - 1 is the abscissa of point B
The abscissa x = 1 and the ordinate y = 1 of point a, i.e. a (1,1)
AC = √{(xA-xC)^2+(yA-yC)^2} = √{(1-(-2))^2+(1-0)^2} = √10
When the coordinate of point C (- 1,1), the linear equation of BC is Y-1 = - 1 (x - (- 1)), that is, y = - X
Substituting y = - x into x ^ 2 + 3Y ^ 2-4 = 0, x ^ 2 + 3x ^ 2-4 = 0, x = ± 1
X = - 1 is the abscissa of point C
The abscissa x = 1 and the ordinate y = - 1 of point B, i.e. B (1, - 1)
The equation of AB is Y - (- 1) = 1 * (x-1), that is, y = X-2
Substituting y = X-2 into x ^ 2 + 3Y ^ 2-4 = 0 results in x ^ 2 + 3 (X-2) ^ 2-4 = 0,4 (x-1) (X-2) = 0, x = 1, or 2
X = 1 is the abscissa of point B
The abscissa x = 2 and the ordinate y = 2-2 = 0 of point a, i.e. a (2,0)
AC = √{(xA-xC)^2+(yA-yC)^2} = √{(2-(-1))^2+(0-1)^2} = √10
In both cases, the AC length is equal
The equation of the line where AB is located:
When the coordinate of point C is (- 2,0), y = X
When the coordinate of point C (- 1,1), y = X-2
Given the inscribed triangle ABC of the ellipse x ^ 2 / 4 + y ^ 2 / 3 = 1, the focus is on the side BC, and a moves on the ellipse, try to find the locus of the center of gravity of the triangle ABC
x²/4+y²/3=1
Because it is an inscribed triangle ABC and the focus is on BC, B and C are the endpoints of the major axis of the ellipse
a²=4
A=2
The coordinates of points B and C are (- 2,0), (2,0) respectively
Let the coordinates of point a be (2cosa, √ 3sina)
The G coordinate of the center of gravity is (x, y)
x=(-2+2+2cosa)/3,
y=(0+0+√3sina)/3
therefore
cosa=3x/2
sina=√3y
cos²a+sin²a=9x²/4+3y²
9x²/4+3y²=1
This is the trajectory equation
It's an ellipse
Proof: hyperbola x ^ 2-y ^ 2 = the product of the distance between any point P and two focal points on a ^ 2 is equal to the square of the distance between P and the center of the hyperbola (a > 0)
Let the coordinates of point p be (x, y)
Then the distance from P to the origin is √ (x ^ 2-y ^ 2) = √ (2x ^ 2-A ^ 2)
So the square of the distance from P to the origin is 2x ^ 2-A ^ 2
By simplifying the hyperbolic equation, we get x ^ 2 / A ^ 2-y ^ 2 / A ^ 2 = 1
According to the formula of intersection radius of hyperbola, the product of two intersection radii is
(ex-a)(ex+a)=(ex)^2-a^2
Because C ^ 2 = a ^ 2 + A ^ 2 = 2A ^ 2, so C = (√ 2) a
e=c/a=√2
So the product of two intersection radii is 2x ^ 2-A ^ 2
So the distance from P to the origin = the product of two intersecting radii
Well, I didn't do it. I can only wish you good luck
If we know sin (x + π / 4) = - 3 / 5, then the value of sin2x is equal to? Please, thank you
sin(x+π÷4) =sinXcosπ/4+sinπ/4sinX =√2/2(sinX+cosX) =-3/5 1/2(sinX+cosX)^2=9/25 1+2sinXcosX=18/25 2sinXcosX=-7/25 sin2x =2sinXcosX=-7/25
It is known that the positive proportional function y = KX (k is not equal to 0) and the linear function y = - x + 8
1. If the images of the first-order function and the positive scale function intersect at point (4, m), find m and K
2. Under what conditions does K satisfy, the intersection of the images of the above two functions must be in the first quadrant?
(1) When x = 4, y = m, M = 4K, M = - 4 + 8, M = 4, k = 1 (2) the positive proportional function y = KX (K ≠ 0) intersects the first-order function y = - x + 8, so the intersection of KX = - x + 8 (K + 1) x = 8, x = 8 / (K + 1), y = KX = 8K / (K + 1) must be in the first quadrant x > 0, Y > 0 8 / (K + 1) > 0, and 8K / (K + 1) > 0. The common solution set is k > 0, K satisfies k > 0
1. Just substitute the intersection point, k = 1, M = 4
2, the intersection point is in the first quadrant, M > 0
Point (4, m) is on y = - x + 8, so m = - 4 + 8 = 4
On K y = 4, so k = 4, k = 4
2：
y=kx
y=-x+8
The continuous solution is x = 8 / (K + 1)
y=8k/(k+1)
The intersection point must be in the first quadrant, indicating that 8 / (K + 1) > 0
8K / (K + 1) >... Expansion
Point (4, m) is on y = - x + 8, so m = - 4 + 8 = 4
(4, m) = (4, 4) on y = KX, so 4 = k * 4, k = 1
2：
y=kx
y=-x+8
The continuous solution is x = 8 / (K + 1)
y=8k/(k+1)
The intersection point must be in the first quadrant, indicating that 8 / (K + 1) > 0
8k/(k+1)>0
The results show that K + 1 > 0 and 8K / (K + 1) > 0
So k > - 1 and k > 0
Comprehensive answer: k > 0
Given the square of the equiaxed hyperbola X - the square of y = the square of a and the point P above it, we prove that the product of the distances from P to its two focal points is equal to the distance from P to the center of the hyperbola
Let the left focus be F1, the right focus be F2, and the center of hyperbola be o (origin of coordinate axis), then a = a, B = a, C = √ 2A
In △ pf1f2, OP is the middle line of F1F2
PF1^2＋PF2^2＝2OP^2＋2OF1^2＝2OP^2＋4A^2 ①
From the definition of hyperbola:
|PF1－PF2|＝2A
(PF1－PF2)^2＝4A^2
PF1^2＋PF2^2－2PF1·PF2＝4A^2 ②
Substitute (1) for (2)
2OP^2＋4A^2－2PF1·PF2＝4A^2
The conclusion is as follows
PF1*PF2=OP^2
Given sin (x + π 4) = 513, then the value of sin2x is equal to ()
A. 120169B. 119169C. −120169D. -119169
Method 1: ∵ sin (x + π 4) = 22 (SiNx + cosx) = - 513, ∵ sin (x + π 4) = - 513, ∵ sin 2x = - Cos2 (x + π 4) = - [1-2sin2 (x + π 4)] = - 119169; method 2: ∵ sin (x + π 4) = - 513, ∵ sin 2x = - Cos2 (x + π 4) = - [1-2sin2 (x + π 4)] = - 119169
If the scale function y = KX (k is not equal to 0) passes through the point (- 1,2), the resolution of the positive scale function is y=
This is my homework tomorrow. Come on
Thank you
It's better to be more detailed
Passing point (- 1,2)
When x = - 1, y = 2
Then 2 = - K
k=-2
So y = - 2x
M
x=a²+2a+1+3
=(a+1)²+3>=3
So m is a set of real numbers greater than or equal to 3
N
y=b²-4b+4+2
=(b-2)²+2>=2
So n is a set of real numbers greater than or equal to 2
It is obvious that n is in the range of M
So m is a proper subset of n
It is proved that the product of the distances from any point P to two focal points on the hyperbola x ^ 2-y ^ 2 = R ^ 2 is equal to the square of the distance from P to the center of the hyperbola
Let the left focus be F1, the right focus be F2, and the center of hyperbola be o (origin of coordinate axis), then a = R, B = R, C = root sign (2) r in △ pf1f2, OP is the middle line of F1F2. According to the middle line theorem, Pf1 ^ 2 + PF2 ^ 2 = 2op ^ 2 + 2of1 ^ 2 = 2op ^ 2 + 4R ^ 2