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《数学通报》2006年第8期“数学问题1628”为:直线l:mx+ny=1与椭圆xa22+yb22=1(a0,b0,a≠b)交于P、Q两点,O为椭圆中心.求证:∠POQ=2π的充要条件是m12+n12=a12+b12.1解法探究证法1联立方程组,消去y得:(a2n2+b2m2)x2-2a2mn2x+a2m2(n2-b2)=0.设P(x1,y1),Q(x2,y2),得x1+x2=2a2mn
Mathematical Problem 1628 is: the straight line l: mx + ny = 1 and the ellipse xa22 + yb22 = 1 (a0, b0, a ≠ b) Elliptical center. Proof: ∠ POQ = 2π necessary and sufficient conditions is m12 + n12 = a12 + b12.1 Solution Exploring the Symmetry 1 Simultaneous Equations, elimination of y: (a2n2 + b2m2) x2-2a2mn2x + a2m2 (n2- b2) = 0. Let P (x1, y1), Q (x2, y2) be x1 + x2 = 2a2mn