∴,即ac=1.

(2)连结PD,交x轴于E,直线PD必为抛物线的对称轴,连结AD、BD,

图代13-3-22

∴                            .

.

∵                             a>0,x2>x1

∴                     .

.

又                                ED=OC=c,

∴                               .

(3)设∠PAB=β,

∵P点的坐标为,又∵a>0,

∴在Rt△PAE中,.

∴                          .

∴                  tgβ=tgα. ∴β=α.∴∠PAE=∠ADE.

∵                           ∠ADE+∠DAE=90°

∴PA和⊙D相切.

  • 答案
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