Question

数学已知椭圆c：x2/a2+y2/b2=1（a＞b＞0）的离心率为√6/3且经过点（3/2，1/2）

求椭圆c的方程

Answer 1

e=c/a=√6/3
c^2/a^2=6/9=2/3
c^2=2/3a^2
b^2=a^2-c^2=a^2-2/3a^2=a^2/3
x^2/a^2+y^2/b^2=1
b^2x^2+a^2y^2=a^2b^2
过(3/2,1/2)
b^2(3/2)^2+a^2(1/2)^2=a^2b^2
9b^2+a^2=4a^2b^2
9a^2/3+a^2=4a^2*a^2/3
4a^2=4a^4/3
a^4/3-a^2=0
a^2(a^2/3-1)=0
a^2=0(舍去)
a^2/3-1=0
a^2=3
b^2=a^2/3=3/3=1
椭圆C方程：x^2/3+y^2=1

追问

过点p（0，2）且斜角为2兀/3的直线交椭圆于A,B两点'求△AOB面积

Answer 2

e = c/a = √6/3 = == a^2 = 3b^2

x^2 + 3y^2 = 3b^2 = = b^2 = 1, a^2= 3

x^2/3 + y^2 = 1

Answer 3

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