Given the eccentricity of ellipse C E = √ 6 / 3, a Quasilinear equation is x = 3 √ 2 / 2 Let P satisfy: OP vector = om vector + on vector, where m and N are the points on the ellipse, and the product of the slopes of OM and on is - 1 / 3. Question: are there two fixed points a and B, so that PA + Pb is the fixed value? If so, find the coordinates of a and B; if not, explain the reason

Given the eccentricity of ellipse C E = √ 6 / 3, a Quasilinear equation is x = 3 √ 2 / 2 Let P satisfy: OP vector = om vector + on vector, where m and N are the points on the ellipse, and the product of the slopes of OM and on is - 1 / 3. Question: are there two fixed points a and B, so that PA + Pb is the fixed value? If so, find the coordinates of a and B; if not, explain the reason

From the eccentricity of ellipse C E = √ 6 / 3, a Quasilinear equation is x = 3 √ 2 / 2
The results are as follows
a=√3,b=1,c=√2
∴x²/3-y²=1
To solve the system of equations - ellipse and the linear equation passing through the origin y = kx
{x²/3-y²=1
{y = KX, we can get x = ± √ (3 / 1 + 3K & # 178;), y = ± K √ (3 / 1 + 3K & # 178;)
Let the point coordinates of M and n be (x1, Y1), (X2, Y2), and the point coordinates of p be (x0, Y0)
Where x0 = X1 + X2, Y0 = Y1 + Y2
M. Where the slope of the line passing through OM is set to K, the slope of the line passing through on is - 1 / 3K
So we can get the coordinate equations of x1, Y1, X2, Y2 with respect to K - K is the only unknown
∴X1=±√(3/1+3k²）,Y1=±k√(3/1+3k²）、X2=±3k√1/(1+3k²）,Y2=±√1/(1+3k²）
x0=±（3k+√3）√1/(1+3k²）,y0==±（√3k-1）√1/(1+3k²）
Square x0 and Y0 respectively
x0²=3+6√3k/(1+3k²）,y0²=1-2√3k/(1+3k²）
Observation easy to get x0 & # 178; / 3 + Y0 & # 178; = 2
We obtain that x0 and Y0 are the points on the hyperbola, and the hyperbolic equation is: x0 & # 178 / 6 + Y0 & # 178 / 2 = 1
So there are two fixed points a and B such that PA + Pb is a fixed value
A=（2√2,0）,B=（-2√2,0）

It is known that the ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) passes through the point (2,0), and the eccentricity is √ 3 / 2,1
2. A1 and A2 are the left and right vertices of ellipse C. the line L: x = 2 √ 2 intersects with X axis at point D. point P is the moving point of ellipse C which is different from A1 and A2. The line a1p and A2p intersect the line L at two points E and f respectively. It is proved that de * DF is constant value 1

1. C / a = √ 3 / 2, C = √ 3 / 2a, B ^ 2 = a ^ 2-C ^ 2 = 1 / 4A ^ 2,
Substituting (2,0) into the equation of ellipse C, it is easy to get a ^ 2 = 4, B ^ 2 = 1
So the equation is: x ^ 2 / 4 + y ^ 2 = 1
2. Prove: let P (X., Y.), K (a1p) = y. / (X. + 2),
So the equation of a1p is: y = y. / (X. + 2) * (x + 2)
When x = 2 √ 2, y = y. / (X. + 2) * (2 √ 2 + 2), that is, de = y. / (X. + 2) * (2 √ 2 + 2),
Similarly, the equation of A2p is: y = y. / (X. - 2) * (2 √ 2-2), │ DF │ = y. / (2-x.) * (2 √ 2-2) (because X. is less than 2),
So, de * DF = 4Y. ^ 2 / (4-x. ^ 2), and P (X., Y.) is on the ellipse x ^ 2 / 4 + y ^ 2 = 1,
So X. ^ 2 + 4Y. ^ 2 = 4, that is, 4Y. ^ 2 = 4-x. ^ 2, so │ de │ * │ DF = 1

Given that ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 has the same eccentricity as ellipse x ^ 2 / 4 + y ^ 2 / 8 = 1, the equation of ellipse C may be ()
A,x^2/8+ y^2/4=m^2 (m≠0) B,x^2/16 + y^2/64=1 C,x^2/8 +y^2/2=1
D. None of the above is possible
Excuse me, how to solve this problem? I remember what parameter should be set to limit the value of a and B~

If the two ellipses have the same eccentricity, then a / b of them are equal,
Because eccentricity e = C / A,
And C = (a * a + b * b) ^ 0.5,
So e = (1 + B / a) ^ 0.5
For this question, a / b = 2 / 2 √ 2 = √ 2 / 2
So what we are looking for is the equation satisfying a / B in the options. It's easy to know that we should choose D

Given that the ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) passes through the point (1,3 / 2), and E = 1 / 2, solve the elliptic equation