22. 解：（1）?PA为圆的切线,??PAD??ABD,?AC 平分
?DAB,??BAC??CAD,?PA为圆的切线,?PA2?PD?PB,?PQ2?PD?PB.
（2）
??PAD??PBA,?PAPB129?,?PB??8,?PA2?PD?PB,?PD?,
3ADAB82915?AQ?DQ?PA?PD?3??.
88?x?3cos??x2y2?1,?射线??与曲23. 解：（1）?C1的参数方程??C1的普通方程?394?y?2sin?线C2交于点D?4,????3?2x?4?y?16. ,的直角坐标方程?C??2?2（2）曲线C1坐标方程为
?2cos2?9??2sin2?4?1,??2?36, 224cos??9sin???12?362,??2224cos??9sin?3636?, 22??????4sin??9cos?4cos2?????9sin2????2?2???4cos2??9sin2?4sin2??9cos2?13?2?2???. ?1?23636361124. 解：（1）由x?1?3?8得?8?x?1?3?8,??11?x?1?5,得不等式的解集为
?x|?4?x?6?.
（2）?任意x1?R,都有x2?R,使得f?x1??g?x2?成立,?y|y?f?x??y|y?g?x?, 又
????f?x??2x?5a?2x?1??2x?5a???2x?1??5a?1,g?x??x?1?3?3,
3424???5a?1?3,解得a?或a??,?实数a的取值范围是?a|a?或a???.
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