论文部分内容阅读
题目在平面直角坐标系xOy中,已知椭圆C的中心在原点,焦点在x轴上,短轴长为2,离心率为(2~(1/2))/2.(Ⅰ)求椭圆C的方程;(Ⅱ)A,B为椭圆C上满足△AOB的面积为(6~(1/2))/2的任意两点,E为线段AB的中点,射线OE交椭圆C于点P.设(?)=t(?),求实数t的值.解(Ⅰ)略.椭圆C的方程为(x~(1/2))/2+y~2=1.
In Cartesian coordinate system xOy, it is known that the center of ellipse C is at the origin, the focal point is on the x-axis, the minor axis is 2 and the eccentricity is (2 ~ (1/2)) / 2. (I) (2) A and B are any two points with an area of (6 1/2 (1/2)) / 2 on the oval C that satisfies Δ AOB, E is the midpoint of the line segment AB, and the ray OE intersects the ellipse C at Point P. Let (?) = T (?), Find the value of real number t. Solution (I) Slightly. The equation of elliptic C is (x ~ (1/2)) / 2 + y ~ 2 = 1.