The ellipse knows 2 focal points. How to find the equation of the ellipse?

The ellipse knows 2 focal points. How to find the equation of the ellipse?

Can't ask. Only know C
sin(a+b)=4/5,tanb=4/3,0
Because 0
Given two focal points (- 3,0) (3,0) and ellipse passing (3,16 / 5), the equation of ellipse is solved
(3,0) F2 (3,0) F1 (3,0) F1 (3,0) F2 (3,0) (3,0) F2 (3,0) (3,0) F2 (3,0) (3,0) (3,0) F2 (3,0) (3,0) F2 (3,0) (3,0) F2 (3,0) (3,0) F2 (3,0) (3,0) F2 (3,0) focuses on the x-axis, and C = 3 ellipseover P (3,16 / 6 / 5) (3,16 / 5) | PF2; (PF2 | = 16 / 5 | PF2 | PF2 | = 16 / 5 | PF2 | (16 / 5 | PF2 | PF2 | (16 / 5 | PF2; (16 / 5 | PF2 | PF2 | usingellipse to define 2
If we know sin α = 1 / 2 plus cos α, then α belongs to the group of 0,2, and the value of sin α - 4 of Cos2 α is 0
Because sin (a - π / 4) = √ 2 (Sina COSA) / 2cos2a = (COSA) ^ 2 - (Sina) ^ 2 = (COSA + Sina) (COSA Sina), so the original formula = - √ 2 / [2 (COSA + Sina)] and because Sina = 0.5 + cosa, so (Sina COSA) ^ 2 = 1 / 4, so 2sinacosa = 3 / 4, so (Sina + COSA) ^ 2 = 7 / 4, because a ∈ (0, π /
It is known that the focus of the ellipse is F1 (0, √ 2) F2 (0, √ 2) and passes through the point (1,3 √ 2 / 2) to find the equation of the ellipse
c=√2
x²/b²+y²/a²=1
Then x & # 178; / (A & # 178; - 2) + Y & # 178; / A & # 178; = 1
1/(a²-2)+9/2a²=1
a²=6
So x & # 178 / 4 + Y & # 178 / 6 = 1
Given sin (α + quarter) = four fifths, and quarter < α < three fourths, find the value of cos α
Given the Quasilinear x = 4 of the ellipse, the corresponding focus is f (2,0) and the eccentricity is 1 / 2
It's not a standard equation. You can do that. The eccentricity is 1 / 2, so a = 2C,
The equation is (x-n) ^ 2 / 4C ^ 2 + y ^ 2 / 3C ^ 2 = 1
The focus is f (2,0), N + C = 2
Quasilinear x = 4 = a ^ 2 / C + n
The solution is C = 2 / 3, a = 4 / 3, n = 4 / 3
The centrifugal ratio is 1 / 2 = C / A
Guide line x = 4 = a ^ 2 / C
The solution is C = 1, a = 2, B ^ 2 = 3
So the elliptic equation is: x ^ 2 / 4 + y ^ 2 / 3 = 1
x=4=a^2/c
1/2＝c/a
So C = 1, a = 2, B ^ 2 = 3
Because the focus is f (2, 0)
^ y (2 / 3) = 1 / 4
The equation is (x-n) ^ 2 / 4C ^ 2 + y ^ 2 / 3C ^ 2 = 1
The focus is f (2,0), N + C = 2
Quasilinear x = 4 = a ^ 2 / C + n
The solution is C = 2 / 3, a = 4 / 3, n = 4 / 3. You can get the equation
It is known that sincos is one fourth of α
sinacosa=1/8
2sinacosa=1/4
(cosa-sina)²=sin²a+cos²a-2sinacosa=1-1/4=3/4
π / 4 ＜ a ＜ π / 2
Here, Sina > cosa
Cosa Sina < 0
Then cosa Sina = - √ 3 / 2
If the two focuses of the ellipse are (- 2,0) and (2,0), and the ellipse passes through the point (5 / 2,3 / 2), then the equation of the ellipse
Let x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1
The results are as follows:
a^2-b^2=2^2````````````````````````1
25b^2/4+9a^2/4=a^2b^2`````````2
The results of simultaneous 1 and 2 are as follows
a^2=10,b^2=6
So: the equation of ellipse x ^ 2 / 10 + y ^ 2 / 6 = 1
According to the elliptic equation: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1, the two focuses of the ellipse are (- 2,0) and (2,0), then a = 2
Replace point (5 / 2,3 / 2), a = 2 with x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 to get
25 / 16 + 9 / 4B ^ 2 = 1, B ^ 2 = 61 / 16, a = 2 is replaced by x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1
x^2/4+16y^2/ 61＝1
Given sin θ 2 + cos θ 2 = 233, then the value of sin θ is______ The value of Cos2 θ is______ .
Analysis: from sin θ 2 + cos θ 2 = 233, 1 + sin θ = 43, sin θ = 13, Cos2 θ = 1-2, sin 2 θ = 1-2 · 19 = 79