Let the center of the ellipse be at the origin o, the focus be on the x-axis, the eccentricity be √ 2 / 2, and the sum of the distances between a point P and two focus points on the ellipse be equal to √ 6. Let the center of the ellipse be at the origin o, the focus be on the x-axis, the eccentricity be √ 2 / 2, and the sum of the distances between a point P and two focus points on the ellipse be equal to √ 6. (1) if the straight line x + y + M = 0 intersects the ellipse at two points a and B, and OA ⊥ ob, find the value of M?

Let the center of the ellipse be at the origin o, the focus be on the x-axis, the eccentricity be √ 2 / 2, and the sum of the distances between a point P and two focus points on the ellipse be equal to √ 6. Let the center of the ellipse be at the origin o, the focus be on the x-axis, the eccentricity be √ 2 / 2, and the sum of the distances between a point P and two focus points on the ellipse be equal to √ 6. (1) if the straight line x + y + M = 0 intersects the ellipse at two points a and B, and OA ⊥ ob, find the value of M?

From the meaning 2A = √ 6, C / a = √ 2 / 2,
So a = √ 6 / 2, C = √ 3 / 2, B & # 178; = A & # 178; - C & # 178; = 3 / 4,
The elliptic equation is X & # 178; / (3 / 2) + Y & # 178; / (3 / 4) = 1, that is, 2x & # 178; + 4Y & # 178; = 3,
From the straight line x + y + M = 0, y = - x-m 2,
Substituting 2 into 1, we get 6x & # 178; + 8mx + 4m & # 178; - 3 = 0,
Let a (x1, Y1), B (X2, Y2)
Then X1 + x2 = - 4m / 3, x1x2 = (4m & # 178; - 3) / 6
y1y2=(-x1-m)(-x2-m)=x1x2+m(x1+x2)+m²=(2m²-3)/6,
Because OA ⊥ ob, so x1x2 + y1y2 = 0, that is (4m & # 178; - 3) / 6 + (2m & # 178; - 3) / 6 = 0,
So M & # 178; = 1, that is, M = 1 or M = - 1

Let the center of the ellipse go back to the origin o, the focus go back to the x-axis, the eccentricity is √ 6 / 3, the sum of the distances from a point P on the ellipse to two focuses is equal to 6, and let the left and right focuses of the ellipse be F1 and F2
(1) Find the standard equation of the ellipse
(2) If Pf1 is perpendicular to PF2, calculate the area of triangle f1pf2 and the coordinates of P at this time

x^2/a^2+y^2/b^2=1
The sum of the distances from P to the two focal points is equal to 6,
So 2A = 6
a=3
e=c/a=√6/3
c=√6
So B ^ 2 = a ^ 2-C ^ 2 = 3
x^2/9+y^2/3=1
F1(-√6,0),F2(√6,0)
F1F2=2√6
P(m,n)
Pf1 vertical PF2
So Pf1 ^ 2 + PF2 ^ 2 = F1F2 ^ 2
(m+√6)^2+n^2+(m-√6)^2+n^2=24
2m^2+2n^2=12
m^2+n^2=6
P is on the ellipse
So m ^ 2 / 9 + n ^ 2 / 3 = 1
m^2+3n^2=9
So n ^ 2 = 3 / 2, m ^ 2 = 9 / 2
|n|=√6/2
So the area is equal to | F1F2 | * | n | / 2 = 3
P has four solutions (3 √ 2 / 2, √ 6 / 2)
(-3√2/2,√6/2)
(3√2/2,-√6/2)
(-3√2/2,-√6/2)

There is a point P (3, y) in the ellipse with the center at the origin and two focal points on the x-axis. If the distance between the point P and the two focal points is 6.5 and 3.5 respectively, the elliptic equation is solved

According to the meaning of the title, let the focus coordinates be F1 (- C, 0), F2 (C, 0) (c > 0)
|PF1| = 6.5; |PF2| =3.5 ==> 2a =|PF1| + |PF2| = 10 ==> a=5；
|PF1|² = (3+c)² + y² =6.5²；---- (1)
|PF1|² = (3-c)² + y² =3.5²； ---- (2)
(1) 2
12c = 30 ==> c = 5/2；
So B & # 178; = A & # 178; - C & # 178; = 75 / 4
The elliptic equation is as follows
x²/25 + y²/(75/4) = 1；

The center of the ellipse is at the origin, the focus is on the Y axis, the focal length is 4, and the eccentricity is two-thirds

The focal length is 4, which means C = 2
Eccentricity = C / A, so a = 3
B = root a ^ 2-C ^ 2 = root 5
The focus is on the y-axis again
The equation comes out
x^2/5+y^2/9=1.