一个高一数学问题 高一数学难题
\u4e00\u4e2a\u9ad8\u4e00\u6570\u5b66\u95ee\u9898\u4ee4y=x²\uff0c\u4e14-1\u2264x\u22643
\u6839\u636e\u629b\u7269\u7ebf\u7684\u56fe\u50cf\uff0cy\u7684\u6700\u5c0f\u503c\u662f0
\u22340\u2264y\u22649
\u5373\uff1a0\u2264x²\u22649
y=(3sinx-3)/(2cosx+10)
=3/2(sinx-1)/(cosx-(-5))
\u7531\u4e8esinx^2+cosx^2=1
\u6240\u4ee5(sinx-1)/(cosx-(-5))
\u53ef\u770b\u6210\u5706x^2+y^2=1\u4e0a\u7684\u70b9\u4e0e(-5,1)\u8fde\u7ebf\u7684\u659c\u7387
\u663e\u7136\u8bbe\u8fde\u7ebf\u4e3ay-1=k(x+5)
\u7136\u540e\u624d\u7528d-r\u6cd5\u7b97\u51fak\u7684\u6700\u503c\uff08\u76f8\u5207\u65f6\u53d6\u5230\uff09
\u7136\u540e\u518d\u6c42\u51fa3/2k\u5c31\u662f\u6240\u6c42
\u6216\u5219\u53e6\u4e00\u79cd
y=(3sinx-3)/(2cosx+10)
2ycosx+1oy=3sinx-3
3sinx-2ycosx=10y+3
\u6839\u636e\u4e09\u89d2\u51fd\u6570\u7684\u6709\u754c\u6027
-\u6839\u53f7\uff084y^2+9)<=3sinx-2ycosx<=\u6839\u53f7\uff084y^2+9)
\u6240\u4ee5-\u6839\u53f7\uff084y^2+9\uff09<10y+3<=\u6839\u53f7\uff084y^2+9\uff09
\u6240\u4ee5(10y+3)^2<=4y^2+9
100y^2+60y+9<=4y^2+9
96y^2+60y<=0
8y^2+5y<=0
\u6240\u4ee5-5/8=<y<=0
\u6240\u4ee5y\u7684\u6700\u503c\u4e3a-5/8\u548c0
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