一:若O和F点分别是椭圆x^2/4+y^2/3=1的中心和左焦点,点P为椭圆上的任意一点,则向量OPX向量FP的最大值是

1个回答

  • 1、当P点在右顶点时二向量积有最大值,

    c=√(4-3)=1,

    OP•FP=|a+c|*|a|*cos0°=|(2+1)|*2=6.

    2、c^2=a^2-b^2=1,c=1,

    直线方程为:y=2(x-1),2x-y-2=0,

    原点至直线距离d=|0-0-2|/√(4+1)=2/√5,

    x^2/5+(2x-2)^2/4=1,

    3x^2-5x=0,

    x1=0,y1=-2,

    x2=5/3,y2=4/3,

    |AB|=√[(x1-x2)^2+(y1-y2)^2]=5√5/3,

    S△AOB=|AB|*d/2=5/3.

    以BA为X轴,BA中点的垂线为Y轴,建立直角坐标系,

    则B(-c,0),A(c,0),

    |BA|=2c,根据椭圆定义,|CA|+|CB|=2a,

    |CA|/|CB|+1=2a/|CB|,

    tanB+1=2a/|CB|,

    3/4+1=2a/|CB|,

    e=c/a,a=c/e,

    secB=√(1+(tanB)^2)=5/4,

    cosB=4/5,

    7/4=2c/(e*|CB|=2/(e*CB/c)=1/(e*|CB|/2c)=1/(e*cosB)=1/(4e/5),

    5/(4e)=7/4,

    e=5/7,

    离心率e=5/7.

    4、 设椭圆方程为:x^2/a^2+y^2/b^2=1,

    左右焦点坐标F1(-c,0),F2(c,0),

    P(x0,y0),

    (x0+c)^2+y0^2=48,(1)

    (x0-c)^2+y0^2=12,(2)

    (1)-(2)式,

    4cx0=36,

    x0=9/c,

    根据三角形角平分线比例线段的性质,

    |PF1|/|PF2|=|F1Q|/|QF2|,

    |F1Q|=1+c,|QF2|=c-1,

    4√3/(2√3)=(1+c)/(c-1),

    c=3,

    x0=9/3=3,

    y0=±2√3,

    b^2=a^2-c^2=a^2-9,

    椭圆方程为:x^2/a^2+y^2/(a^2-9)=1,

    将P点坐标代入椭圆方程求出a,

    a^4-30a^2+81=0,

    (a^2-27)(a^2-3)=0,

    a^2=27,a^2=3,

    因c=3,故舍去a^2=3,

    a^2=27,

    则椭圆方程为:x^2/27+y^2/18=1.

    5、设椭圆方程为x^2/a^2+y^2/b^2=1,(a>b>0),暂设焦点在X轴,

    e=c/a=√3/2,c=√3a/2,b^2=a^2-c^2=a^2/4,

    椭圆方程为:x^2/a^2+y^2/(a^2/4)=1,

    y=-x/2-4,代入椭圆方程,

    2x^2+16x+64-a^2=0,

    根据韦达定理,

    x1+x2=-8,x1*x2=32-a^2/2,

    根据弦长公式,

    |PQ|=√(1+k^2)(x1-x2)^2

    =√(1+1/4)[(x1+x2)^2-4x1*x2]

    =(1/2)√[5*(64-128+2a^2)]

    =(1/2)√(10a^2-320)

    =√10,

    a=6,

    b=3,

    椭圆方程为:x^2/36+y^2/9=1,

    若焦点在Y轴,一样可做.

    6、设A(x1,y1),B(x2,y2),AB直线斜率k,

    x1^2/16+y1^2/4=1,(1)

    x2^2/16+y2^2/4=1,(2)

    (1)-(2)式,

    (x1^2-x2^2)/16+(y1^2-y2^2)/4=0,

    1/4+[(y1-y2)/(x1-x2)]/{[(y1+y2)/2]/[(x1+x2)/2]}=0,

    k=( y1-y2)/(x1-x2),

    (y1+y2)/2和(x1+x2)/2为M点纵、横坐标,

    1/4+k*(1/2)=0,

    k=-1/2,

    则直线方程为:(y-1)/(x-2)=-1/2,

    即x+2y-4=0.

    故存在这样的直线.