高考數(shù)學(xué)專題復(fù)習(xí)導(dǎo)練測 第九章 解析幾何階段測試（十三）課件 理 新人教A版.ppt

《高考數(shù)學(xué)專題復(fù)習(xí)導(dǎo)練測 第九章 解析幾何階段測試（十三）課件 理 新人教A版.ppt》由會員分享，可在線閱讀，更多相關(guān)《高考數(shù)學(xué)專題復(fù)習(xí)導(dǎo)練測 第九章 解析幾何階段測試（十三）課件 理 新人教A版.ppt（28頁珍藏版）》請在裝配圖網(wǎng)上搜索。

數(shù)學(xué) A（理）,45分鐘階段測試(十三),第九章 平面解析幾何,,,2,3,4,5,6,7,8,9,10,1,,,2,3,4,5,6,7,8,9,10,,1,,,2,3,4,5,6,7,8,9,10,,1,答案 C,2.設(shè)雙曲線的一個焦點(diǎn)為F，虛軸的一個端點(diǎn)為B，如果直線FB與該雙曲線的一條漸近線垂直，那么此雙曲線的離心率為( ),,,3,4,5,6,7,8,9,10,1,,2,,,3,4,5,6,7,8,9,10,1,,2,答案 D,3.已知拋物線關(guān)于x軸對稱，它的頂點(diǎn)在坐標(biāo)原點(diǎn)O，并且經(jīng)過點(diǎn)M(2，y0).若點(diǎn)M到該拋物線焦點(diǎn)的距離為3，則|OM|等于( ),,,2,4,5,6,7,,8,9,10,1,3,,,2,4,5,6,7,8,9,10,1,,3,答案 B,,,2,3,5,6,7,8,9,10,1,,4,,,2,3,5,6,7,8,9,10,1,,4,,,2,3,5,6,7,8,9,10,1,,4,即橢圓C上有4個點(diǎn)是“★”點(diǎn).,答案 B,5.已知拋物線y2＝2px (p0)，過其焦點(diǎn)且斜率為1的直線交拋物線于A、B兩點(diǎn)，若線段AB的中點(diǎn)的縱坐標(biāo)為2，則該拋物線的準(zhǔn)線方程為( ) A.x＝1 B.x＝－1 C.x＝2 D.x＝－2,,,2,3,4,6,7,8,9,10,1,,5,,,2,3,4,6,7,8,9,10,1,,5,①－②得(y1－y2)(y1＋y2)＝2p(x1－x2).,又線段AB的中點(diǎn)的縱坐標(biāo)為2，即y1＋y2＝4，,直線AB的斜率為1，故2p＝4，p＝2，,答案 B,二、填空題 6.已知拋物線y2＝2px (p0)的準(zhǔn)線與圓x2＋y2－6x－7＝0相切，則p的值為__.,,,2,3,4,5,7,8,9,10,1,,6,圓x2＋y2－6x－7＝0可化為(x－3)2＋y2＝16，,2,,,2,3,4,5,6,8,9,10,1,,7,,,2,3,4,5,6,8,9,10,1,,7,可得4c2＋36＝4a2， 即a2－c2＝9，故有b＝3. 答案 3,,,2,3,4,5,6,7,9,10,1,,8,,,2,3,4,5,6,7,9,10,1,,8,,,2,3,4,5,6,7,9,10,1,,8,答案 ＝,,,2,3,4,5,6,7,8,10,1,,9,,,2,3,4,5,6,7,8,10,1,,9,,,2,3,4,5,6,7,8,10,1,,9,(2)設(shè)直線l過橢圓的焦點(diǎn)且與圓C相切，求直線l的方程.,,,2,3,4,5,6,7,8,10,1,,9,,,2,3,4,5,6,7,8,9,1,,10,解 由已知，得x＝4不合題意，,設(shè)直線l的方程為y＝k(x－4)，,由已知，得拋物線C的焦點(diǎn)坐標(biāo)為(1,0)，,,,2,3,4,5,6,7,8,9,1,,10,,,2,3,4,5,6,7,8,9,1,,10,(2)設(shè)A，B為拋物線上兩點(diǎn)，且AB不垂直于x軸，若線段AB的垂直平分線恰過點(diǎn)M，求證：線段AB中點(diǎn)的橫坐標(biāo)為定值.,證明 設(shè)線段AB中點(diǎn)的坐標(biāo)為N(x0，y0),A(x1，y1),B(x2，y2)，,,,2,3,4,5,6,7,8,9,1,,10,,,2,3,4,5,6,7,8,9,1,,10,所以x0＝2，即線段AB中點(diǎn)的橫坐標(biāo)為定值2.,