答案:(11√5)/5
8.圆心在抛物线x2=2y上,与直线2x+2y+3=0相切的圆中,面积最小的圆的方程为 .
解析:圆心在抛物线x2=2y上,设圆心为(x"," 1/2 x^2 ),直线2x+2y+3=0与圆相切,则圆心到直线2x+2y+3=0的距离等于圆的半径,即r=("|" 2x+x^2+3"|" )/√(2^2+2^2 )=("|" x^2+2x+3"|" )/(2√2)=("|(" x+1")" ^2+2"|" )/(2√2)≥2/(2√2)=√2/2,当x=-1时,r最小,即圆的面积最小,此时圆的圆心坐标为("-" 1"," 1/2),故圆的方程为(x+1)2+(y"-" 1/2)^2=1/2.
答案:(x+1)2+(y"-" 1/2)^2=1/2
9.顶点为坐标原点,焦点在x轴上的抛物线截直线y=-2x-1所得的弦长|AB|=5√3,则抛物线的方程为 .
解析:设抛物线的方程为y2=2mx(m≠0),点A的坐标为(x1,y1),点B的坐标为(x2,y2),{■(y^2=2mx"," @y="-" 2x"-" 1)┤⇒4x2+(4-2m)x+1=0⇒{■(x_1+x_2=(m"-" 2)/2 "," @x_1 x_2=1/4 "," )┤
∴5√3=√5·√(((m"-" 2)/2)^2 "-" 4×1/4)⇒m=10或-6,
∴y2=20x或y2=-12x.
答案:y2=20x或y2=-12x
10.已知点P(2,0),点Q在曲线C:y2=2x上.
(1)若点Q在第一象限内,且|PQ|=2,求点Q的坐标;
(2)求|PQ|的最小值.
解(1)设Q(x0,y0)(x0>0,y0>0),由题意得|PQ|=√("(" x_0 "-" 2")" ^2+y_0^2 )=2,∴(x0-2)2+y_0^2=4.0①
又点Q(x0,y0)在曲线C上,∴y_0^2=2x0.0②
由①②,得x0=y0=2.
∴点Q的坐标为(2,2).
(2)设Q(x0,y0),∵点Q在曲线C上,∴y_0^2=2x0,
∴|PQ|=√("(" x_0 "-" 2")" ^2+y_0^2 )=√("(" x_0 "-" 2")" ^2+2x_0 )=√(x_0^2 "-" 2x_0+4)=√("(" x_0 "-" 1")" ^2+3).