椭圆x2/a2+y2/b2=1(a>b>0)的离心率e=根号3/2,点A(0,3/2)与椭圆上的点的最大距离是根号7求椭

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  • 椭圆x^2/a^2+y^2/b^2=1(a>b>0)的离心率e=√3/2,

    所以c^2/a^2=3/4,a^2=4c^2/3,b^2=a^2-c^2=c^2/3.

    所以椭圆方程是x^2/(4c^2/3)+y^2/(c^2/3)=1,

    点A(0,3/2)与椭圆上的点P((2c/√3)cosu,(c/√3)sinu)的距离

    |AP|=√{[(2c/√3)cosu]^2+[(c/√3)sinu-3/2]^2},

    AP^2=(4c^2/3)(cosu)^2+(c^2/3)(sinu)^2-√3csinu+9/4

    =-c^2*(sinu)^2-√3csinu+4c^2/3+9/4

    =-c^2*[sinu+√3/(2c)]^2+4c^2/3+3的最大值是7,

    1)√3/(2c)1,sinu=-1时AP^2取最大值c^2/3+√3c+9/4=7,

    c0,c=(-3√3+2√21)/2,与c