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第1394题:圆锥曲线切线方程2



本题做为知识拓展,在答案中输入1即可过关. 


选自陈劲松、李世臣发表于《数学教学通讯》上的“经过圆锥曲线外一点的切线方程公式”一文.


过椭圆外一定点 (x0,y0)(x_0,y_0)椭圆 x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 所引的切线方程为


(x0xa2+y0yb21)2(\dfrac{x_0x}{a^2}+\dfrac{y_0y}{b^2}-1)^2 =(x02a2+y02b21)=(\dfrac{x_0^2}{a^2}+\dfrac{y_0^2}{b^2}-1) (x2a2+y2b21)(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}-1)


过双曲线外一定点 (x0,y0)(x_0,y_0)双曲线 x2a2y2b2=1\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1  所引的切线方程为


(x0xa2y0yb21)2(\dfrac{x_0x}{a^2}-\dfrac{y_0y}{b^2}-1)^2 =(x02a2y02b21)=(\dfrac{x_0^2}{a^2}-\dfrac{y_0^2}{b^2}-1) (x2a2y2b21)(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}-1)


过抛物线外一定点 (x0,y0)(x_0,y_0)抛物线 y2=2pxy^2=2px  所引的切线方程为


[y0yp(x0+x)]2[y_0y-p(x_0+x)]^2 =(y022px0)=(y_0^2-2px_0)(y22px) (y^2-2px)

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