
(1)求雙曲線C的方程;
(2)過(guò)點(diǎn)P的動(dòng)直線l與C的左、右兩支分別交于A,B兩點(diǎn),若點(diǎn)M在線段AB上,滿足eq \f(|AP|,|AM|)=eq \f(|BP|,|BM|),證明:點(diǎn)M在定直線上.
思路分析
?利用離心率和eq \(PF1,\s\up6(—→))·eq \(PF2,\s\up6(—→))=6求方程
?設(shè)直線方程y-1=k?x-3?并聯(lián)立
?利用比例關(guān)系eq \f(|AP|,|AM|)=eq \f(|BP|,|BM|)列式
?將根與系數(shù)的關(guān)系代入化簡(jiǎn)
?消去參數(shù)得點(diǎn)在定直線上
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[子題1] (2023·信陽(yáng)模擬)已知橢圓C:eq \f(x2,4)+eq \f(y2,2)=1的左、右頂點(diǎn)分別為A1,A2,過(guò)點(diǎn)D(1,0)的直線l與橢圓C交于異于A1,A2的M,N兩點(diǎn).若直線A1M與直線A2N交于點(diǎn)P,證明:點(diǎn)P在定直線上,并求出該定直線的方程.
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[子題2] (2023·岳陽(yáng)模擬)已知雙曲線C:x2-eq \f(y2,3)=1,P為雙曲線的右頂點(diǎn),設(shè)直線l不經(jīng)過(guò)P點(diǎn)且與C相交于A,B兩點(diǎn),若直線PA與直線PB的斜率之和為-1.證明:直線l恒過(guò)定點(diǎn).
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規(guī)律方法 動(dòng)線過(guò)定點(diǎn)問(wèn)題的兩大類型及解法
(1)動(dòng)直線l過(guò)定點(diǎn)問(wèn)題,解法:設(shè)動(dòng)直線方程(斜率存在)為y=kx+t,由題設(shè)條件將t用k表示為t=mk,得y=k(x+m),故動(dòng)直線過(guò)定點(diǎn)(-m,0).
(2)動(dòng)曲線C過(guò)定點(diǎn)問(wèn)題,解法:引入?yún)⒆兞拷⑶€C的方程,再根據(jù)其對(duì)參變量恒成立,令其系數(shù)等于零,得出定點(diǎn).
1.(2023·襄陽(yáng)模擬)過(guò)拋物線x2=2py(p>0)內(nèi)部一點(diǎn)P(m,n)作任意兩條直線AB,CD,如圖所示,連接AC,BD并延長(zhǎng)交于點(diǎn)Q,當(dāng)P為焦點(diǎn)并且AB⊥CD時(shí),四邊形ACBD面積的最小值為32.
(1)求拋物線的方程;
(2)若點(diǎn)P(1,1),證明:點(diǎn)Q在定直線上運(yùn)動(dòng),并求出定直線方程.
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2.(2023·全國(guó)乙卷)已知橢圓C:eq \f(y2,a2)+eq \f(x2,b2)=1(a>b>0)的離心率是eq \f(\r(5),3),點(diǎn)A(-2,0)在C上.
(1)求C的方程;
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(2)過(guò)點(diǎn)(-2,3)的直線交C于P,Q兩點(diǎn),直線AP,AQ與y軸的交點(diǎn)分別為M,N,證明:線段MN的中點(diǎn)為定點(diǎn).
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這是一份專題六 第4講 母題突破3 定值問(wèn)題2024年高考數(shù)學(xué),共2頁(yè)。試卷主要包含了已知雙曲線C,已知橢圓C等內(nèi)容,歡迎下載使用。
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