1.å·²ç¥åC:x²+y²-2x+4y-4=0æ¯å¦åå¨æçä½1çç´çº¿mï¼ä½¿ä»¥m被åCæªå¾ç弦AB为ç´å¾çåè¿åç¹ï¼è¥åå¨ï¼æ±åºç´çº¿mçæ¹ç¨ï¼è¥ä¸åå¨ï¼è¯´æçç±ã
解ï¼è®¾ç´çº¿mçæ¹ç¨ä¸ºx=y+b.................(1)
代å
¥åCçæ¹ç¨ï¼(y+b)²+y²-2(y+b)+4y-4=0
åç®å¾2y²+2(b+1)y+b²-2b-4=0..............(2)
设交ç¹A(x₁,y₁); B (x₂,y₂).y₁,y₂æ¯æ¹ç¨ï¼2ï¼çæ ¹ï¼æ
ï¼
y₁+y₂=-(b+1)
y₁y₂=(b²-2b-4)/2
x₁+x₂=y₁+b+y₂+b=(y₁+y₂)+2b=-(b+1)+2b=b+1
x₁x₂=(y₁+b)(y₂+b)=y₁y₂+b(y₁+y₂)+b²=(b²-2b-4)/2-b(b+1)+b²=(b²-4b-4)/2
弦é¿âABâ=â [(x₁-x₂)²+(y₁-y₂)²]=â[(x₁+x₂)²+(y₁+y₂)²-4(x₁x₂+y₁y₂)]
=â{[b+1)²+(b+1)²-4[(b²-4b-4)/2+(b²-2b-4)/2]}
=â(-2b²+16b+18)
设弦ABä¸ç¹çåæ 为(m,n),åm=(x₁+x₂)/2=(b+1)/2, n=(y₁+y₂)/2=-(b+1)/2.
ABä¸ç¹ï¼m,n)å°åç¹çè·ç¦»d=â[(b+1)²/2]=âABâ/2=(1/2)â(-2b²+16b+18)
æ
æ(b+1)²/2=(1/4)(-2b²+16b+18)
åç®å¾ b²-3b-4=(b-4)(b+1)=0,æ
b=4æ-1.
äºæ¯å¾ç´çº¿mçæ¹ç¨ä¸ºï¼x=y+4, æx=y-1,å³
y-x+4=0æy-x-1=0为ææ±ã
2.å·²ç¥åCï¼x²+ï¼y-1ï¼²=5ï¼ç´çº¿l:mx-y+1-m=0
æ±ï¼1ï¼æ±è¯mâRï¼ç´çº¿lä¸åCæ»æ两个ä¸å交ç¹ï¼
ï¼2ï¼è®¾ä¸åC交ä¸ä¸åç两个ç¹ABï¼è¥|AB|=æ ¹å·17ï¼æ±lçå¾æè§ï¼
ï¼3ï¼æ±å¼¦ABçä¸ç¹Mç轨迹æ¹ç¨
解ï¼(1)åå¿C(0ï¼1ï¼å°ç´çº¿Lçè·ç¦»d=â0-1+1-mâ/â(m²+1)=âmâ/â(m²+1)<1
å³0<d<1<â5ï¼ï¼åå¾ï¼å¯¹ä»»ä½mé½æç«ï¼æ
ç´çº¿Lä¸åæ»æ两个交ç¹ã
ï¼2ï¼ãå°ç´çº¿æ¹ç¨y=mx+1-m代å
¥åçæ¹ç¨å¹¶åç®å¾ï¼
(1+m²)x²-2m²x+m²-5=0................(1)
设交ç¹A(x₁y₁);B(x₂,y₂), x₁,x₂æ¯æ¹ç¨ï¼1ï¼çæ ¹ï¼æ
ï¼
x₁+x=2m²/(1+m²)
x₁x₂=(m²-5)/(1+m²)
y₁+y₂=(mx₁+1-m)+(mx₂+1-m)=m(x₁+x₂)-2m+2=2m³/(1+m²)-2m+2=(m-1)²/(1+m²)
y₁y₂=(mx₁+1-m)(mx₂+1-m)=m²(x₁x₂)+m(1-m)(x₁+x₂)+(1-m)²
=m²(m²-5)/(1+m²)+2m³(1-m)/(1+m²)+(1-m)²=(1-3m)(1+m)/(1+m²)
âABâ=â[(x₁+x₂)²+(y₁+y₂)²-4(x₁x₂+y₁y₂)](代å
¥ä¸é¢çç»æï¼ï¼â17
ç±æ¤è§£åºm,è¿ä¸ªmå°±æ¯ç´çº¿Lçæçã
ï¼3ï¼è®¾å¼¦ABä¸ç¹çåæ 为ï¼x,y),é£ä¹
x=(x₁+x₂)/2=m²/(1+m²)
y=(y₁+y₂)/2=(m-1)²/2(1+m²)
两å¼èç«æ¶å»ååém,å³å¾ABä¸ç¹ç轨迹æ¹ç¨ã
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