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圓錐曲線的斜率問題

2022-08-12 17:20 作者:因戀相思久  | 我要投稿

如圖,已知圓O : x%5E2%2B%20y%5E2%20%3D4,點(diǎn)B(1,0),以線段AB為直徑的圓內(nèi)切于圓0,點(diǎn)A的集合記為曲線C.

(1)求曲線C的方程;

(2)已知直線l%EF%BC%9Ax%3D4,Q(1%2C%5Cfrac%7B3%7D%7B2%7D%20),過點(diǎn)B的直線l_%7B1%7D%20與C交于M , N 兩點(diǎn),與直線l交于點(diǎn)K,記QM%2CQN%2CQK?的斜率分別為k_%7B1%7D%20%EF%BC%8Ck_%7B2%7D%20%EF%BC%8Ck_%7B3%7D%20問:%5Cfrac%7Bk_%7B1%7D-%20k_%7B2%7D%20%7D%7Bk_%7B2%7D-%20k_%7B3%7D%20%7D%20是否為定值?若是,給出證明,并求出定值:若不是,說明理由.

(1)設(shè)AB的中點(diǎn)為P ,切點(diǎn)為Q,連接OP,PQ,取B關(guān)于y軸的對稱點(diǎn)D,連接AD,則%5Cvert%20AD%20%5Cvert%20%3D2%5Cvert%20OP%5Cvert%20

%5Cvert%20AB%20%5Cvert%2B%20%5Cvert%20AD%20%5Cvert%3D%202%5Cvert%20OP%20%5Cvert%2B2%20%5Cvert%20PB%20%5Cvert%20%3D2%5Cvert%20OP%20%5Cvert%2B%202%5Cvert%20PQ%5Cvert%20

%3D2(%5Cvert%20OP%20%5Cvert%2B%20%5Cvert%20PQ%20%5Cvert)%20%3D4%EF%BC%9E%5Cvert%20BD%20%5Cvert%20%3D2

所以點(diǎn)A的軌跡是以B,D為焦點(diǎn),長軸長為4的橢圓其中a =2, c = 1, b =%5Csqrt%7B3%7D%20.則曲線C的方程為

%5Cfrac%7Bx%5E2%7D%7B4%7D%20%2B%5Cfrac%7By%5E2%20%7D%7B3%7D%20%3D1

(2)設(shè)%E7%9B%B4%E7%BA%BFMN%EF%BC%9Any%2B1%3Dx,M(x_%7B1%7D%EF%BC%8C%20y_%7B1%7D%20)%EF%BC%8CN(x_%7B2%7D%EF%BC%8C%20y_%7B2%7D%20)

因此有K%EF%BC%884%EF%BC%8C%5Cfrac%7B3%7D%7Bn%7D%20%EF%BC%89,k_%7B3%7D%20%3D%5Cfrac%7B%5Cfrac%7B3%7D%7Bn%7D-%5Cfrac%7B3%7D%7B2%7D%20%20%7D%7B4-1%7D%20%3D%5Cfrac%7B1%7D%7Bn%7D%20-%5Cfrac%7B1%7D%7B2%7D%20

k_%7B1%7D%20%3D%5Cfrac%7By_%7B1%7D%20-%5Cfrac%7B3%7D%7B2%7D%20%7D%7Bx_%7B1%7D%20-1%7D%20%3D%5Cfrac%7By_%7B1%7D%20-%5Cfrac%7B3%7D%7B2%7D%7D%7Bny_%7B1%7D%20%7D%20

k_%7B2%7D%20%3D%5Cfrac%7By_%7B2%7D%20-%5Cfrac%7B3%7D%7B2%7D%20%7D%7Bx_%7B2%7D%20-1%7D%20%3D%5Cfrac%7By_%7B2%7D%20-%5Cfrac%7B3%7D%7B2%7D%7D%7Bny_%7B2%7D%20%7D%20

k_%7B1%7D%2B%20k_%7B2%7D%20%3D%5Cfrac%7B2y_%7B1%7D%20y_%7B2%7D-%5Cfrac%7B3%7D%7B2%7D%20%EF%BC%88y_%7B1%7D%2B%20y_%7B2%7D%20%EF%BC%89%20%7D%7Bny_%7B1%7D%20y_%7B2%7D%20%7D%3D%5Cfrac%7B2%7D%7Bn%7D-%20%20%5Cfrac%7B3(y_%7B1%7D%20%2By_%7B2%7D%20)%7D%7B2ny_%7B1%7D%20y_%7B2%7D%20%7D%20

3x%5E2%2B%204y%5E2%3D12%E4%B8%8Eny%2B1%3Dx%E8%81%94%E7%AB%8B

%EF%BC%883n%5E2%2B4%EF%BC%89y%5E2%2B6ny-9%3D0

y_%7B1%7D%2B%20y_%7B2%7D%20%3D-%5Cfrac%7B6n%7D%7B3n%5E2%2B4%20%7D%20,y_%7B1%7D%20y_%7B2%7D%20%3D-%5Cfrac%7B9%7D%7B3n%5E2%2B4%20%7D%20

因此3%EF%BC%88y_%7B1%7D%20%2By_%7B2%7D%20%EF%BC%89%3D2ny_%7B1%7D%20y_%7B2%7D%20

%20%20%5Cfrac%7B3(y_%7B1%7D%20%2By_%7B2%7D%20)%7D%7B2ny_%7B1%7D%20y_%7B2%7D%20%7D%20%3D1

所以k_%7B1%7D%2B%20k_%7B2%7D%20%3D%5Cfrac%7B2%7D%7Bn%7D%20-1%3D2k_%7B3%7D%20

k_%7B1%7D%20-k_%7B2%7D%20%3D-2(k_%7B2%7D%20-k_%7B3%7D%20)

所以%5Cfrac%7Bk_%7B1%7D-%20k_%7B2%7D%20%7D%7Bk_%7B2%7D-%20k_%7B3%7D%20%7D%20%3D-2為定值

下面進(jìn)行系統(tǒng)化:

已知橢圓%5Cfrac%7Bx%5E2%20%7D%7Ba%5E2%20%7D%2B%20%5Cfrac%7By%5E2%20%7D%7Bb%5E2%20%7D%20%3D1%EF%BC%88a%EF%BC%9Eb%EF%BC%9E0%EF%BC%89,直線l_%7B1%7D%20%3Ax%3D%5Cfrac%7Ba%5E2%20%7D%7Bc%7D%20F%EF%BC%88c%EF%BC%8C0%EF%BC%89%EF%BC%8CQ%EF%BC%88c%EF%BC%8Ct%EF%BC%89t%5Cneq%200,過點(diǎn)F的直線l_%7B1%7D%20與橢圓交于M , N 兩點(diǎn),與直線l_%7B1%7D%20交于點(diǎn)K,記QM%2CQN%2CQK?的斜率分別為k_%7B1%7D%20%EF%BC%8Ck_%7B2%7D%20%EF%BC%8Ck_%7B3%7D%20

設(shè)%E7%9B%B4%E7%BA%BFMN%EF%BC%9Any%2Bc%3Dx%EF%BC%8CM(x_%7B1%7D%EF%BC%8C%20y_%7B1%7D%20)%EF%BC%8CN(x_%7B2%7D%EF%BC%8C%20y_%7B2%7D%20)

因此有K%EF%BC%88%5Cfrac%7Ba%5E2%20%7D%7Bc%7D%20%EF%BC%8C%5Cfrac%7Bb%5E2%20%20%7D%7Bnc%7D%20%EF%BC%89k_%7B3%7D%20%3D%5Cfrac%7B%5Cfrac%7Bb%5E2%20%7D%7Bnc%7D%20-t%7D%7B%5Cfrac%7Ba%5E2%20%7D%7Bc%7D-c%20%7D%20%3D%5Cfrac%7B%5Cfrac%7Bb%5E2%20%7D%7Bnc%7D%20-t%7D%7B%5Cfrac%7Bb%5E2%20%7D%7Bc%7D%20%20%7D%3D%5Cfrac%7B1%7D%7Bn%7D%20-%5Cfrac%7Btc%7D%7Bb%5E2%20%7D%20

k_%7B1%7D%20%3D%5Cfrac%7By_%7B1%7D%20-t%7D%7Bx_%7B1%7D%20-c%7D%20%3D%5Cfrac%7By_%7B1%7D%20-t%7D%7Bny_%7B1%7D%20%7D%20

k_%7B2%7D%20%3D%5Cfrac%7By_%7B2%7D%20-t%20%7D%7Bx_%7B2%7D%20-c%7D%20%3D%5Cfrac%7By_%7B2%7D%20-t%7D%7Bny_%7B2%7D%20%7D%20

k_%7B1%7D%2B%20k_%7B2%7D%20%3D%5Cfrac%7B2y_%7B1%7D%20y_%7B2%7D-t%EF%BC%88y_%7B1%7D%2B%20y_%7B2%7D%EF%BC%89%20%7D%7Bny_%7B1%7D%20y_%7B2%7D%7D%20%3D%5Cfrac%7B2%7D%7Bn%7D%20-%5Cfrac%7Bt%EF%BC%88y_%7B1%7D%2B%20y_%7B2%7D%EF%BC%89%7D%7Bny_%7B1%7D%20y_%7B2%7D%7D%20

b%5E2%20x%5E2%2B%20a%5E2%20y%5E2%3Da%5E2%20b%5E2%20%E4%B8%8Eny%2Bc%3Dx%E8%81%94%E7%AB%8B

%EF%BC%88n%5E2%20b%5E2%2Ba%5E2%20%EF%BC%89y%5E2%2B2ncb%5E2y-b%5E4%3D0

y_%7B1%7D%2B%20y_%7B2%7D%20%3D-%5Cfrac%7B2ncb%5E2%20%7D%7Ba%5E2%2B%20n%5E2%20b%5E2%20%7D%20,y_%7B1%7D%20y_%7B2%7D%20%3D-%5Cfrac%7Bb%5E4%20%7D%7Ba%5E2%2B%20n%5E2%20b%5E2%20%7D%20

因此b%5E2%20%EF%BC%88y_%7B1%7D%20%2By_%7B2%7D%20%EF%BC%89%3D2ncy_%7B1%7D%20y_%7B2%7D%20

%5Cfrac%7B(y_%7B1%7D%20%2By_%7B2%7D%20)%7D%7Bny_%7B1%7D%20y_%7B2%7D%20%7D%20%3D%5Cfrac%7B2c%7D%7Bb%5E2%20%7D%20

所以k_%7B1%7D%2B%20k_%7B2%7D%20%3D%5Cfrac%7B2%7D%7Bn%7D%20-%5Cfrac%7Bt%EF%BC%88y_%7B1%7D%2B%20y_%7B2%7D%EF%BC%89%7D%7Bny_%7B1%7D%20y_%7B2%7D%7D%20%3D%5Cfrac%7B2%7D%7Bn%7D%20-%5Cfrac%7B2tc%7D%7Bb%5E2%20%7D%20%3D2k_%7B3%7D%20

最后,善于利用結(jié)論來猜測結(jié)果去推導(dǎo)是很重要的方法

圓錐曲線的斜率問題的評論 (共 條)

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