Through the coordinate origin, the line L and the ellipse (x-3) ^ 2 / 6 + y ^ 2 / 2 = 1 intersect at two points a and B. if the circle with diameter AB just passes through the left focus F of the ellipse Find the inclination angle of a straight line

Through the coordinate origin, the line L and the ellipse (x-3) ^ 2 / 6 + y ^ 2 / 2 = 1 intersect at two points a and B. if the circle with diameter AB just passes through the left focus F of the ellipse Find the inclination angle of a straight line

X & # 178 / / 6 + Y & # 178 / / 2 = 1: C = √ (A & # 178; - B & # 178;) = √ (6-2) = 2F (- 2,0) (x-3) & # 178 / / 6 + Y & # 178 / / 2 = 1, which is obtained by shifting 3 units to the right of X & # 178 / / 6 + Y & # 178 / / 2 = 1, f (1,0) let the slope of l be K, and the equation is y = KX substituting (x-3) & # 178 / / 6 +

Let the coordinates of point a be (1,1 / 2), the intersection ellipse x ^ 2 / 4 y ^ 2 = 1 passing through the origin o at points B and C, and find the maximum area of △ ABC
We have calculated s = 1 / 2|k-1 / 2| / √ 1 + 4k2, and then?

S^2=(k^2-k+1/4)/[4(1+4k^2)]
(4+16k^2)S^2=k^2-k+1/4,
(16S^2-1)k^2+k+4S^2-1/4=0,k∈R,
∴△=1-(16S^2-1)^2>=0,
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Ellipse C: X & # 178 / 3 + Y & # 178; = 1, the line L intersects the ellipse at two points a and B, if l passes through point Q (0,2), find the maximum area of △ AOB

Elliptic equation x ^ 2 / 3 + y ^ 2 = 1
The distance between the line and the origin is fixed √ 3 / 2, and the line and the ellipse intersect at two points a and B
Finding the maximum area of △ AOB is equivalent to finding the maximum distance of ab
The height is a fixed value, which is equivalent to making a tangent on a circle with a radius of √ 3 / 2, and finding the maximum distance between the tangent and the intersection of the ellipse
If the semicircle is √ 3 / 2, the equation is: x ^ 2 + y ^ 2 = 3 / 4
Obviously, when the tangent is perpendicular to the X axis, the distance of AB is the largest, and the tangent is x = √ 3 / 2
Substituting x = √ 3 / 2 into the elliptic equation, we get (√ 3 / 2) ^ 2 / 3 + y ^ 2 = 1
The solution is y = ± √ 3 / 2, and the maximum value of | ab | = 2 * √ 3 / 2 = √ 3
The maximum area of ∧ AOB is s ∧ AOB = 1 / 2 * √ 3 / 2 * √ 3 = 3 / 4
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If the point m (3,0), the ellipse x24 + y2 = 1 and the line y = K (x + 3) intersect at points a and B, then the perimeter of △ ABM is______ ．

In the ellipse x24 + y2 = 1, a = 2, B = 1, C = 3, m (3,0) is the right focus of the ellipse, the line y = K (x + 3) is the left focus of the ellipse, and the circumference of ∧ ABM is 4A = 8, so the answer is: 8