(西南師大附中高2009級第三次月考)已知圓C經(jīng)過P.Q兩點(diǎn).且在y軸上截得的線段長為.半徑小于5. (1)求直線PQ與圓C的方程. (2)若直線l∥PQ.且l與圓C交于點(diǎn)A.B..求直線l的方程. 解:(1) PQ為···························································· 2分 C在PQ的中垂線即y = x – 1上································ 3分 設(shè)C(n.n – 1).則······································· 4分 由題意.有······································································· 5分 ∴ ∴ n = 1或5.r 2 = 13或37(舍)······················ 7分 ∴圓C為············································································· 8分 解法二: 設(shè)所求圓的方程為 由已知得解得 當(dāng)時.,當(dāng)時.(舍) ∴ 所求圓的方程為 (2) 設(shè)l為···················································································· 9分 由.得································ 10分 設(shè)A(x1.y1).B(x2.y2).則················ 11分 ∵ . ∴ ···················································· 12分 ∴ ∴ ∴ m = 3或 – 4(均滿足) ∴ l為···························································· 14分 查看更多

 

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(08年東北師大附中文)(12分)

已知等差數(shù)列,公差,且 ,成等比數(shù)列,

(Ⅰ)求的通項公式;

(Ⅱ)設(shè)的前項和為,求證:

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.師大附中青華校區(qū)從三樓半到四樓共有11級臺階,某老師一步可以上一級、二級或三

    級臺階,若步上完,共有                    種不同方法.(用數(shù)字作答)

 

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(08年哈師大附中文) 過拋物線的焦點(diǎn)作一條直線與拋物線相交于兩點(diǎn),且,則這樣的直線有

   A.一條    B.兩條    C.三條    D.不存在

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(08年哈師大附中文)為了解某校高三學(xué)生的視力情況,隨機(jī)地抽查了該校100名高三學(xué)生的視力情況,得到頻率分布直方圖(如右圖),由于不慎將部分?jǐn)?shù)據(jù)丟失,但知道前4組頻數(shù)成等比數(shù)列,后6組的頻數(shù)成等差數(shù)列,若最大頻率為a,視力在4.6到5.0之間的學(xué)生數(shù)為b,則a、b的值分別為

A.0.27和78        B.0.09和75           C.0.22和58             D.0.27和75

 

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(08年東北師大附中文) 已知,則下列不等式成立的是

(A)

(B)

(C)

(D)

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