Given the ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), A1 and A2 are the left and right vertices of the ellipse. Let F1 be the focus of the ellipse, It is proved that if and only if the point P on C is at the left and right vertices, Pf1 takes the maximum and minimum

Given the ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), A1 and A2 are the left and right vertices of the ellipse. Let F1 be the focus of the ellipse, It is proved that if and only if the point P on C is at the left and right vertices, Pf1 takes the maximum and minimum

Let the coordinates of point p be (m, n)
Then - a ≤ m ≤ a
Then ipf1i = a + em
When m = - A, ipf1imin = a + e (- a) = a-c (C ^ 2 = a ^ - B ^ 2)
When m = a, ipf1imax = a + EA = a + C
Get proof

It is known that A1 (- 5,0), A2 (5,0) are the two vertices of the ellipse, F1 (- 4,0), F2 (4,0) are the two focuses of the ellipse
(1) Write out the equation of ellipse and its quasilinear equation;
(2) Through any point K on the line oa2 which is different from O and A2, make a vertical line of oa2, intersect ellipse at P and P1, and intersect straight line a1p and a2p1 at point M. prove that point m is on hyperbola x ^ 2 / 25-y ^ 2 / 9 = 1

1) A = 5, C = 4, so B2 = 9
So the elliptic equation is x2 / 25 + Y2 / 9 = 1
The Quasilinear equation is x = A2 / C = 25 / 16

The left and right focus of the ellipse e: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 are F1 and F2 respectively. The left and right vertex are A1 and A2 respectively. T (1,3 / 2) is the point on the ellipse, and T 2 is perpendicular to the X axis. The equation of the ellipse e is obtained

The x-axis means that the abscissa of F2 is the same as that of T, so if F2 is (1,0), then C = 1, F1 (- 1,0) [symmetric relation with F2], F1F2 = 2, and TF2 = 3 / 2
Connect TF1
Then TF1 = √ (TF2 ^ 2 + F1F2 ^ 2) = √ (2 ^ 2 + (3 / 2) ^ 2) = 5 / 2
TF1+TF2=3/2+5/2=4=2a
a=2 b^2=a^2-c^2=2^2-1=3
The equation of ellipse is x ^ 2 / 4 + y ^ 2 / 3 = 1

The vertex of ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > 0, b > 0) is A1, A2, B1, B2, the focus is F1, F2, AB is root 7, parallelogram a1b1a2b2 = parallelogram
Two times the area of a1b1a2b2, find the equation of ellipse C?

It is known that the left vertex A1, the right focus F2 and the point P of the ellipse x ^ 2 / 4 + y ^ 2 / 3 = 1 are the points on the ellipse
When vector PA1 * vector PF2 takes the minimum value, the value of vector PA1 + vector PF2 | is

Let P (a, b)
Because: x ^ 2 / 4 + y ^ 2 / 3 = 1, the left vertex is A1, and the right focus is F2
Then: A1 (- 2,0) F2 (1,0)
Then: vector PA1 = (- 2-A, - b)
Vector PF2 = (1-A, - b)
Because: P is on the ellipse
A ^ 2 / 4 + B ^ 2 / 3 = 1
Then: B ^ 2 = (12-3a ^ 2) / 4 (- 2)

It is known that a and B are two points on the ellipse x ^ 2 / A ^ 2 + 15y ^ 2 / 9A ^ 2 = 1, and F2 is the right focus of the ellipse. If the absolute value of af2 + the absolute value of BF2 = 8 / 5A, the distance from the midpoint of AB to the alignment line of the ellipse is 3 / 2, the elliptic equation can be solved

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It is known that the two focal points of the ellipse are F1, F2 triangle af1f2 is the midpoint of equilateral triangle AF1, and B is just on the ellipse, then the eccentricity of the ellipse is

If BF2 is connected, then BF2 ⊥ AF1, let AF1 = 4, then BF1 = 2, BF2 = 2 √ 3, so a = √ 3 + 1, C = F1F2 / 2 = 2, so eccentricity e = 2 / (√ 3 + 1) = √ 3-1

It is known that F1 and F2 are the left and right focal points of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), a is a point in the first quadrant of the ellipse, point B is also on the ellipse, and satisfies the vector OA + vector ob = 0 (o is the coordinate origin), vector af2 * vector F1F2 = 0, and the eccentricity of the ellipse is equal to / 2
(1) Solving the equation of line ab
(2) If the area of triangle abf2 is equal to 4 √ 2, the equation of ellipse is obtained
(3) Under the condition of (2), is there a point m on the ellipse such that the area of the triangle mAb is equal to 8 √ 3? If there is, find the coordinate of the point m; if not, explain the reason

Let a (x1, Y1) B (X2, Y2) since OA + ob = 0, then there will be X1 + x2 = 0, Y1 + y2 = 0, so the line AB must be symmetrical about the origin, that is, the line is y = kxaf2 * F1F2 = 0, af2 is perpendicular to the X axis, you should find a point a, a (C, B ^ 2 / a) find the slope, and finally use a, B to substitute the most

It is known that F1 and F2 are the left and right focus of the ellipse x * x / A * a + y * y / b * b = 1 (a > b > 0), and a is the point in the first quadrant of the ellipse
(1) The linear AB equation (2) if the area of triangle abf2 is equal to the square root of four, the elliptic equation
(3) Under the condition of (2), whether there is a point m on the ellipse such that the area of the triangle mAb is equal to three eighths of the root sign, exists, and the coordinates of the point m are obtained, but does not exist,

Point B is also on the ellipse, and satisfies the vector OA + vector ob = 0, af2 · F1F2 = 0, the eccentricity is equal to half of the root sign. The process of solving two problems is shown in the figure

It is known that F1 and F2 are the left and right focal points of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), and a is the point in the first quadrant of the ellipse, if AF1 * af2 = 0,
The eccentricity of the circle is equal to 2 / 2 of the root sign, and the area of the triangle aof2 is 2 * root sign 2. Find the standard square of the ellipse
AF1 * af2 = 0 is a vector
The answer is x ^ 2 / 16 + y ^ 2 / 8 = 1
The headline above is wrong. Sorry again! Change "AF1 * af2 = 0" to "af2 * F1F2 = 0"

Because af2 * F1F2 = 0, af2 ⊥ F1F2
AF1+AF2=2a
AF1²=AF2²+F1F2²
(AF1+AF2)(AF1-AF2)=4c²
AF1-AF2=2c²/a
AF1+AF2=2a
2AF2=2a-2c²/a
AF2=a-c²/a
Ao is the center line on the side of F1F2
So s △ aof2 = 1 / 2S △ af1f2
S△AF1F2=4√2
1/2F1F2×AF2=4√2
AF2=4√2/c
therefore
a-c²/a=4√2/c
a²-c²=4√2a/c
a²-c²=b²
C / a = √ 2 / 2 put into the above formula
b²=8
a=√2c
a²=2c²
a²=b²+c²
So B & sup2; = C & sup2; = 8
a²=16
So the equation: X & sup2 / 16 + Y & sup2 / 8 = 1