It is known that the eccentricity e of the ellipse x 2A2 + y 2B2 = 1 (a > b > 0) is 63, and the distance between the straight line passing through points a (0, - b) and B (a, 0) and the origin is 32

It is known that the eccentricity e of the ellipse x 2A2 + y 2B2 = 1 (a > b > 0) is 63, and the distance between the straight line passing through points a (0, - b) and B (a, 0) and the origin is 32

The equation of straight line AB is XA + y − B = 1, that is BX ay AB = 0. From the meaning of the title, we get aba2 + B2 = 32, ① ∵ CA = 63, ② A2 = B2 + C2, ③ the solution, ①, ②, ③ gets a = 3, B = 1. The standard equation of ellipse is x23 + y2 = 1

The eccentricity of ellipse (x ^ 2 / A ^ 2) + (y ^ 2 / b ^ 2) = 1 (a is greater than B is greater than 0) e = √ 6 / 3, the distance of straight line passing through point a (0, - b) and B (a, 0) passing through the origin is √ 3 / 2
(1) (2) Let f 1 and F 2 be the left and right focal points of the ellipse, cross the ellipse through F 2 and make a straight line at P and Q, and find the maximum radius r of the inscribed circle of the triangle PQF 1

(1) The slope of the line is B / A, so the equation of the line is (B / a) x-y-b = 0
So the distance from the line to the origin d = | - B | / √ (B / a) ^ 2 + 1 = √ 3 / 2
e^2=c^2/a^2=(a^2-b^2)/a^2=1-b^2/a^2=2/3
So B ^ 2 / A ^ 2 = 1 / 3, substituting into the algebraic expression of D, there is b = 1
So a = √ 3
So the equation of ellipse is x ^ 2 / 3 + y ^ 2 = 1
(2) The radius of the inscribed circle is r = 2S △ C, where s is the area of the triangle and C is the perimeter of the triangle
We know that in an ellipse, the sum of the points on the ellipse and the two focal points of the ellipse is a constant, that is 2A
The perimeter of the triangle in the title is 4a, that is 4 √ 3
Connecting PQ, the equation of the line where the line is located is y = K (x - √ 2) --- passing through the focus F2 (√ 2,0)
The elliptic equation is x ^ 2 / 3 + y ^ 2 = 1, and the deformation is x ^ 2 + 3Y ^ 2 = 3
Simultaneous equations have
x^2=(y-k√2)^2/k^2
Substituting into the deformed elliptic equation
y^2-2√2ky+2k^2+3k^2*y^2=3k^2
(3k^2+1)y^2-2√2ky-k^2=0
Y1 + Y2 is the sum of the heights of triangle pf1f2 and qf1f2
y1+y2=2√2k/(3k^2+1)
Bottom F1F2 = 2C = 2 √ 2
2 * s triangle pqf2 = 2 * (1 / 2) * 2 √ 2 * 2 √ 2K / (3K ^ 2 + 1) = 8K / (3K ^ 2 + 1)
Perimeter of triangle pqf2 = 4A = 4 √ 3
So r = 2S / C = 2 √ 3K / (9K ^ 2 + 3)

A question in a reference book has a score of 15
It is known that the eccentricity e of ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is 1 / 2, and the distance from origin o to straight line X / A + Y / b = 1 is d = (2 √ 21) / 7
Problem 1: solving the equation of ellipse
Question 2: cross point m (√ 3,0) to make a straight line intersecting with ellipse C at two points P and Q, and find the maximum area of △ OPQ

D = 1 / √ (1 / A ^ 2 + 1 / b ^ 2) = 2 √ 21 / 71 / A ^ 2 + 1 / b ^ 2 = 7 / 12 11-B ^ 2 / A ^ 2 = e ^ 2 = 1 / 43 / 4A ^ 2-B ^ 2 = 0, a ^ 2 = 4B ^ 2 = 3 in 1, so x ^ 2 / 4 + y ^ 2 / 3 = 1 (2) connect the vertical line equation and elliptic equation to get Y1 + Y2 and Y1 * Y2 of the equation about y, and then find (y1-y2) ^ 2 = 39-27 / (1 + 4K ^ 2) this