The ellipse equation is x 2A2 + y 2B2 = 1 (a > b > 0), a vertex is a (0,2), eccentricity e = 63. (1) the equation for solving ellipse; (2) the line L: y = kx-2 The ellipse equation is x2 / A2 + Y2 / B2 = 1 (a > b > 0), a vertex is a (0,2), and the eccentricity e = root 6 / 3 (1) Find the equation of ellipse; (2) The straight line L: y = kx-2 (K ≠ 0) intersects the ellipse at two different points m, n satisfies MP → = PN →, AP → &; Mn → = 0, find K

1. If a vertex is (0,2), it must be on the top of the y-axis, that is, B = 2. Because e = C / a = radical 6 / 3 and a ^ 2-B ^ 2 = C ^ 2, the above equation can be solved to a = 2 radical 3. So the equation is x ^ 2 / 12 + y ^ 2 / 4 = 1
2. MP → = PN → we can know that point P is the middle point of Mn, and AP → &; Mn → = 0 means AP ⊥ Mn, that is, AP is the vertical line of Mn, that is to say, am = an. Let m and N coordinates be (x1, Y1) and (X2, Y2) respectively. Then am ^ 2 = X1 ^ 2 + (y1-2) ^ 2 = X1 ^ 2 + (kx1-4) ^ 2, similarly, an ^ 2 = x2 ^ 2 + (kx2-4) ^ 2. So x2 ^ 2 + (kx2-4) ^ 2 = X1 ^ 2 + (kx1-4) ^ 2, X1 ^ 2-x2 ^ 2 = (kx2-4) ^ 2 - (kx1-4) ^ 2. The final simplification is X1 + x2 = - K [K (x1 + x2) - 8]. By removing y from the line and ellipse, we can get (3K ^ 2 + 1) x ^ 2-12kx = 0. According to Weida's theorem, we can get X1 + x2 = 12K / (3K ^ 2 + 1), and substitute it into the above formula to get k = ± 1 / 3

As shown in the figure, it is known that the eccentricity of the ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is √ 3 / 2. Take the left vertex t of the ellipse C as the center of the circle to make the circle T:
(x + 2) ^ 2 + y ^ 2 = R (R & gt; 0), let t and C intersect at M and n
(1) Find the equation of ellipse C;
(2) Find the minimum value of the vector TM multiplied by the vector TN, and find the equation of the circle t at this time;
(3) Let p be any point on the ellipse C different from m and N, and the intersection of the straight lines MP and NP with the x-axis at the points R, s and o as the origin of the coordinates, and prove that or OS is the fixed value

As shown in the figure, the equation x & # 178 / A & # 178; + Y & # 178 / B & # 178; = 1 (a > b > 0), a is the left vertex of the ellipse, B and C are on the ellipse, if the quadrilateral oabc is a parallelogram and ∠ OAB = 45 °, then the eccentricity of the ellipse is equal to

Eccentricity of ellipse = √ 6 / 3

The elliptic equation with the same focus as the ellipse 9x2 + 4y2 = 36 and the minor axis length of 45 is______ ．

Ellipse 9x2 + 4y2 = 36, C = 5, the focus of ellipse is the same as that of ellipse 9x2 + 4y2 = 36, the half focal length of ellipse is C = 5, that is, A2-B2 = 5, the length of minor axis is 45, B = 25, a = 5, the standard equation of ellipse is y225 + x220 = 1, so the answer is y225 + x220 = 1

Find the standard equation of ellipse 9x & # 178; + 4Y & # 178; = 36 with the same focus, and the short axis length is 4 root sign 5!

It is known that the center of the ellipse e is at the coordinate origin, the focus is on the coordinate axis, and passes through three points a (- 2,0), B (2,0) and C (1,32). (1) the equation for solving the ellipse e: (2) if point D is any point on the ellipse e different from a and B, f (- 1,0), H (1,0), when the area of the inscribed circle of △ DFH is the largest, the coordinates of the center of the inscribed circle can be obtained

(1) Let the elliptic equation be MX2 + Ny2 = 1 (M & gt; 0, n & gt; We substitute a (- 2,0), B (2,0), C (1,32) into the equation of ellipse e, and get 4m = 1m + 94n = 1, and M = 14, n = 13. The equation of ellipse e x24 + Y23 = 1 (2) | FH | = 2, let the height of △ DFH edge be h, s △ DFH = 12 × 2 × H = h. When point D is at the top vertex of ellipse, the maximum value of H is 3, so the maximum value of s △ DFH is 3. Let the radius of the inscribed circle of △ DFH be r, because the perimeter of △ DFH is a fixed value of 6 So 12R × 6 = s △ DFH, so the maximum value of R is 33. So the coordinates of the center of the inscribed circle are (0, ± 33)

The focal point is on the x-axis, the focal length is 2, and the sum of the distances between a point m and two focal points on the ellipse is 6

The sum of the distances from any point on the ellipse to the two focal points is equal to the length of the major axis of the ellipse, that is, 2a, so a is equal to 3, and because C is equal to 1, the square of B equals the square of a minus the square of C equals 8
So the elliptic equation is: X squared / 9 + X squared / 8 = 1

It is known that the focal point of an ellipse is F 1 (0, - 2). F 2 (0,2). The sum of the distances between the points on the ellipse and the two focal points is 8

The focus is F1 (0, - 2). F2 (0,2)
therefore
2c=4
c=2
The sum of the distances from the point on the ellipse to the two focal points is 8
therefore
2a=8
a=4
therefore
b²=a²-c²=16-4=12
The standard equation is as follows
x²/12+y²/16=1

Let the focal coordinates of the ellipse with the center at the origin be F1 (- 1.0) F2 (1.0) and the distances between a point P and two focal points on the ellipse be 1.3, respectively

Let the focal coordinates of the ellipse with the center at the origin be F1 (- 1.0) F2 (1.0) and the distances between a point P and two focal points on the ellipse be 1 and 3, respectively
According to the definition of ellipse, 2A = 1 + 3 = 4, a = 2, a & sup2; = 4, C = 1, so B & sup2; = 2 & sup2; - 1 & sup2; = 3,
The elliptic equation is X & sup2 / 4 + Y & sup2 / 3 = 1

1. The two focal points of the ellipse are F1 (- 4,0) and F2 (4,0), and the sum of the distances from a point on the ellipse to the two focal points is 12
Equation

Because the focus is on the x-axis, let the equation be X & # 178 / A & # 178; + Y & # 178 / B & # 178; = 1, let a point on the ellipse be p, connecting Pf1 and PF2
Because F1 and F2 are focal coordinates,
So C = 4, C & # 178; = 16
And because | Pf1 | + | PF2 | = 2A = 12,
So a = 6, a & # 178; = 36
Because in the ellipse A & # 178; = B & # 178; + C & # 178;,
So B & # 178; = 36-16 = 20
So the equation of ellipse is X & # 178 / 36 + Y & # 178 / 20 = 1