过点(2,3)作动直线l交椭圆x²/4+y²=1于不同的点P,Q,过P,Q作椭圆的切线,两条切线的交
1个回答
(1)设切点P(x1,y1),Q(x2,y2),则
切线PM:x1x/4+y1y=1,
QM:x2x/4+y2y=1,
它们都过点M(m,n),
∴x1m/4+y1n=1,
x2m/4+y2n=1,
∴直线l:mx/4+ny=1过P,Q,
直线PQ过点(2,3),
∴m/2+3n=1,即m+6n-2=0,
以(x,y)代(m,n)得x+6y-2=0,
M在椭圆x^2/4+y^2=1外,
∴M的轨迹方程是x+6y-2=0(x^2/4+y^2>1).
化简得x+6y-2=0(y3/5).
(2)设M(2-6n,n),则l:(1-3n)x/2+ny=1,y=[1-(1-3n)x/2]/n,
代入椭圆方程得x^2/4+[1-(1-3n)x/2]^2/n^2=1,
n^2x^2+4-4(1-3n)x+(1-3n)^2x^2=4n^2,
整理得(10n^2-6n+1)x^2-4(1-3n)x+4-4n^2=0,
△=16(1-3n)^2-16(10n^2-6n+1)(1-n^2)
=16(10n^4-6n^3-9n^2+6n-1
+9n^2-6n+1)
=16(10n^4-6n^3),
|PQ|=√△/(10n^2-6n+1)*√{1+[(1-3n)/(2n)]^2}
=√[(10n^2-6n)(13n^2-6n+1)]/(10n^2-6n+1),
M到l的距离d1=|(1-3n)^2+n^2-1|/√{[（1-3n)/2]^2+n^2]
=2(10n^2-6n)/√(13n^2-6n+1),
O到l的距离d2=2/√(13n^2-6n+1),
∴S(POQM)=(1/2)|PQ|(d1+d2)=√[(10n^2-6n)=4,
平方得5n^2-3n-8=0,
解得n=-1或8/5,
这时l的方程是2x-y-1=0,或19x-16y+10=0.
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