解析几何大题在高考中常作为压轴难题，分值极大，但复杂的计算往往令考生们望而却步。本文旨在教你如何跳过直线与椭圆联立的复杂计算，透过现象看本质，利用简化版的硬解定理，从而秒杀高考的解析几何大题。

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工具/原料

纸、笔、脑子

方法/步骤

1.根据题设的已知条件，利用待定系数法列出二元二次方程，求出椭圆的方程，并化为标准方程。

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2.直线设为斜截式y=kx+m，将直线与椭圆联立得到如图一元二次方程。注意该式子具有普适性，由高中生学习(gzsxuexige)根据硬解定理简化而来。

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3.通常要验证判别式大于零（因为无论是该经验所给的弦长公式还是韦达定理都是在判别式大于零的情况下才有意义，若题目给出直线与椭圆相交则略去该步，多写不扣分）。

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4.如图所示，直接写出需要的弦长公式或韦达定理。该图可以省去你至少5分钟，而且不会算错，因为你根本就不用算。

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5.恒成立问题的证明可能会与导数，不等式交汇。恒成立问题的证伪只要找到反例即可。存在性问题通常是存在的，方法是提出无关的未知数。

6.最后别忘了写综上所述。

END

注意事项

未验证判别式大于零会酌情扣1—2分

系数的记忆需从b2+a2k2入手，观察四个韦达定理的规律.

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