Given that the focus of the ellipse is F1 (0, - 1) and F2 (0,1), P is a point on the ellipse, and | F1F2 | is the median of | Pf1 | and | PF2 | then the equation of the ellipse is______ .

Given that the focus of the ellipse is F1 (0, - 1) and F2 (0,1), P is a point on the ellipse, and | F1F2 | is the median of | Pf1 | and | PF2 | then the equation of the ellipse is______ .

| f1f1f2 | is the mean of the difference between | Pf1 | and | PF2 | and | PF2 ? 2 | f1f1f2 ???| F1F2 | is the mean of the difference between | Pf1 ? and | PF2 | and | PF2 | and | PF2 |, | 2 | 2 || f1f1f2 | is the mean mean of | Pf1 | Pf1 | Pf1 ? Pf1 | Pf1 ? Pf1 ? Pf1 ? and ? PF2 ? PF2 inthis paper, we discuss the properties of ∵ ellipse ∵ = 2, B2 = a2-c2 = 3, and ∵ ellipse ∵
Let cos (π / 4 + x) = 3 / 5, X ∈ (3 π / 4,7 π / 4), find (sin2x + 2Sin & sup2; x) / (1-tanx)
3π/4
Given the focus of the ellipse F1 (- 1,0), F2 (1,0), P is a point on the ellipse, and | F1F2 | is the median of | Pf1 | and | PF2 | if the point P is in the second quadrant, and ∠ pf1f2 = 120 °, find Tan ∠ f1pf2
According to the topic meaning C = 12f1f2 = Pf1 + pf24c = 2AA = 2C = 2B & sup2; = A & sup2; - C & sup2; = 4-1 = 3 elliptic equation: X & sup2 / / 4 + Y & sup2 / / 3 = 1 let Pf1 = x, then PF2 = 2a-x = 4-xf1f2 = 2, according to cosine theorem cos ∠ pf1f2 = [x & sup2; + 4 - (4-x) & sup2;] / (2 * x * 2) - 1 / 2 = (X & sup2; + 4-16 + 8x-x)
│F1F2│=2
Let PF2 = x, then Pf1 = 4-x
According to the cosine theorem:
x²=（4-x）²+ 4 + 2（4-x）
x=2.8
That is, Pf1 = 2.2
According to cosine theorem cos ∠ f1pf2 =...
Respondent: lcp4347583 | grade 3 | 2011-2-18 22:54
According to the meaning of the title
C=1
2F1F2=PF1+PF2
4c=2a
a=2c=2
b²=a²-c²=4-1=3
Elliptic equation: X & sup2 / 4 + Y & sup2 / 3 = 1
Let Pf1 = x, then PF2 = 2... Expand
Respondent: lcp4347583 | grade 3 | 2011-2-18 22:54
According to the meaning of the title
C=1
2F1F2=PF1+PF2
4c=2a
a=2c=2
b²=a²-c²=4-1=3
Elliptic equation: X & sup2 / 4 + Y & sup2 / 3 = 1
Let Pf1 = x, then PF2 = 2a-x = 4-x
F1F2=2
According to the cosine theorem
cos∠PF1F2=[x²+4-(4-x)²]/(2*x*2)
-1/2=(x²+4-16+8x-x²)/(4x)
8x-12=-2x
10x=12
x=6/5
cos∠F1PF2=[x²+(4-x)²-4]/(2*x*(4-x)]=(36/25+196/25-4)/[2*(6/5)*(14/5)]=11/14
sin∠F1PF2=5√3/14
tan∠F1PF2=[5√3/14]/(11/14)=5√3/11
│F1F2│=2
Let PF2 = x, then Pf1 = 4-x
According to the cosine theorem:
x²=（4-x）²+ 4 + 2（4-x）
x=2.8
That is, Pf1 = 2.2
According to cosine theorem cos ∠ f1pf2 =... Put it away
∫sin2x/sin^4x+cos^4xdx?
∫sin2x/sin^4x+cos^4xdx=∫sin2x/[(sin²x+cos²x)²-2sin²xcos²x]dx=∫sin2x/[1-1/2sin²2x]dx=2∫sin2x/[2-sin²2x]dx=2∫sin2x/(1+cos²2x)dx=-∫1/(1+cos²2x)dcos2x=-a...
The title is ∫ sin2x / (sin ^ 4x + cos ^ 4x) DX. halo
It is known that F1 and F2 are the left and right focal points (1) of the ellipse x ^ 2 / 25 + y ^ 2 / 9 = 1 to find the coordinates of F1F2
(2) If AB is a chord of F1 passing through the focus of the ellipse, find the perimeter of △ abf2
(1) Ellipse x ^ 2 / 25 + y ^ 2 / 9 = 1
a²=25,b²=9,c²=a²-b²=16
∴c=4,
Ψ focus F1 (- 4,0) F2 (4,0)
(2)
∵ a, B are on the ellipse
According to the definition of ellipse:
∴|AF1|+|AF2|=2a=10
|BF1|+|BF2|=2a=10
Add:
|AF1|+|AF2|+|BF1|+|BF2|=20
∵|AF1|+|BF1|=|AB
∴|AB|+|AF2|+|BF2|=20
That is, the perimeter of △ abf2 is 20
(generally, the perimeter of △ abf2 is 4a)
Given sin ^ 4x + cos ^ 4x = 5 / 8, find cos4x?
Because: sin ^ 2x + cos ^ 2x = 1 both sides square at the same time
Sin ^ 4x + cos ^ 4x + 2Sin ^ 2xcos ^ 2x = 1
(2sinxcosx) ^ 2 / 2 = 1-5 / 8
sin(2x)^2=3/4.
So: 2Sin (2x) ^ 2 = 3 / 2. So: 1-2sin (2x) ^ 2 = 1-3 / 2 = - 1 / 2. That is: cos (4x) = - 1 / 2
It is known that the two focuses of ellipse x ^ 2 / A ^ 2 + y ^ 2 / 25 = 1 are F1 and F2, and | F1F2 | = 8, then a=
F1F2 = 2c, so C = 4, because B ^ 2 = 25, so B = 5, and because a ^ 2 = B ^ 2 + C ^ 2, so a = root 41
Can you give me the process
Sin (4 / π - x) = 5 / 13, then sin2x is equal to
sin(4/π-x)=5/13,
cosx-sinx=5√2/13,
sin2x=2sinxcosx=1-(cosx-sinx)²=119/169
On some problems of mathematical focal length, we know that the focal point of an ellipse is F1 (0. - 1), F2 (0.1), P is the point on the ellipse, and F1F2 is the median of Pf1 and PF2
But the distance of F1F2 should be calculated as root 2 with the distance formula, but F1F2 = C and C = 1. How do I understand it
F1F2=2C.
The distance of F1F2 should not be calculated as root 2 with the distance formula
Here C = 1
So f12c = F2
Given sin (x-4 / π) = 3 / 5, then sin 2x=
The known answer is 7 / 25
Sin2x = cos (2x - π / 2) = the square of 1 - sin (x - π / 4)
Note: double angle formula
It should be π / 4