It is known that the focus of the ellipse C is on the x-axis, the eccentricity e = 3 / * 3, and the area of the quadrilateral with four vertex structure s = 2 * 6 It is known that the focus of the ellipse C is on the x-axis, the eccentricity e = 3 / * 3, and the area s of the quadrilateral with four vertex structure is 2 * 6. The standard equation for finding the ellipse C is 〈 * means root 〉

That quadrilateral is a quadrilateral whose diagonals are perpendicular to each other. The quadrilateral area s = half of the product of diagonals. In this problem, 2A * 2B / 2 = 2Ab = 2 √ 6. Therefore, ab = √ 6 and C / a = √ 3 / 3C & # 178; = A & # 178; - B & # 178; triple join can get a & # 178; = 3, B & # 178; = 2. Therefore, the formula is: X & # 178 / 3 + Y -

The two focal points are F1 (- 2,0), F2 (2,0), and passing through P (5 / 2, - 3 / 2) to find the standard equation of ellipse

One focus is F1 (- 2 radical 3,0). The sum of major axis length and minor axis length is only 12

c=2√3 ,2a+2b=12 ,
So a + B = 6,
Because C ^ 2 = a ^ 2-B ^ 2 = (a + b) (a-b) = 12,
So A-B = 2,
So the solution is a = 4, B = 2,
So the standard equation of ellipse is x ^ 2 / 16 + y ^ 2 / 4 = 1

Given that the two focal points of an ellipse are F1 = - 2,0, F2 = (2,0) and point a (0,2) is on the ellipse, then the standard equation of the ellipse is

Solution 1
Defined by ellipse
AF1+AF2=2a
So √ [(0 + 2) ^ 2 + (2-0) ^ 2] + √ [(0-2) ^ 2 + (2-0) ^ 2] = 4 √ 2 = 2A
a=2√2
c=2
b^2=a^2-c^2=2
So x ^ 2 / 8 + y ^ 2 / 4 = 1
Solution 2
The minor axis is the vertical bisector of F1F2
A is right on the vertical bisector of F1F2
So a is the endpoint of the minor axis
So B = 2, C = 2
a^2=b^2+c^2=8
So x ^ 2 / 8 + y ^ 2 / 4 = 1

It is known that the focus of the ellipse is F1 (0, - 3) and F2 (0,3), and it passes through the point (4,0)

c=3
The focus is on the y-axis
So (4,0) is the vertex of the minor axis
b=4
a^2=b^2+c^2=25
x^2/16+y^2/25=1

If we know that the two focal points of an ellipse are F1 (- 1,0) F2 (1,0) and pass through (0, √ 3), then the standard equation of an ellipse is

Let the standard equation be x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1
a^2=b^2+1
3/b^2=1
a^2=4 b^2=3
So the standard equation is x ^ 2 / 4 + y ^ 2 / 3 = 1

It is known that the focus of the ellipse is F1 (0, - 1), F2 (0,1), and the straight line y = 4 is a directrix of the ellipse. The standard equation of the ellipse is solved

It is known that C = 1, y = 4 is a directrix of an ellipse, i.e. a & sup2. / C = 4, then a & sup2; = 4
B & sup2; = A & sup2; - C & sup2; = 4-1 = 3, the focus is on the y-axis,
So the standard equation of ellipse is X & sup2 / 3 + Y & sup2 / 4 = 1

It is known that the focus of ellipse C is F1 (- C, 0) F2 (C, 0), (c > 0) and B = C √ 3, a-c = 2 (1). Find the standard equation of ellipse C (2) and make a straight line through the left focus F1
It is known that the C focus of ellipse is F1 (- C, 0) F2 (C, 0), (c > 0) and B = C √ 3, a-c = 2
(1) Finding the standard equation of ellipse C
(2) Through the left focus F1, make a straight line intersection ellipse C at P and Q, and calculate the perimeter of △ f2pq

(1) A = C + 2, a ^ 2 = B ^ 2 + C ^ 2, the square of (c + 2) = the square of (C √ 30) + the square of C, the solution is C = 2, a = 4, B = 2 √ 3,
The standard equation of ellipse is x ^ 2 / 16 + y ^ 2 / 12 = 1
(2) The perimeter of △ f2pq = (PF2 + Pf1) + (QF2 + QF1) = 4A = 16

Find the elliptic standard equation with focus F1 (0, - 1) F2 (0,1) and passing through point m (3 / 2,1)

2a=MF1+MF2=√(9/4+4)+3/2=4
a=2
a²=4
c²=1
b²=3
So x & # 178 / 3 + Y & # 178 / 4 = 1

An ellipse standard equation with focus F1 (- 2.0) is solved through point B (0. - 4)

Focus (- 2,0)
So C = 2
Passing through point B (0. - 4)
So B = 4
Then a = √ (2 & # 178; + 4 & # 178;) = 2 √ 5
therefore
The elliptic equation is
x²/20+y²/16=1