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这是教材上的一道习题: 求经过两条曲线x~2+y~2+3x-y=0①和3x~2+3y~2+2x+y=0②交点的直线方程。启蒙阶段,可先解交点,后求直线方程: 由①×3-②,可得7x-4y=0③又由①、②联立解之得:x_1=0,y_1=0;x_2=-4/13,y=-7/13。由此得所求的直线方程:7x-4y=0 ④比较③、④,发现由③到④是条回路,于是回头研究式③为所求的道理;若(x_1、y_1)、(x_2、y_2)是两曲线的交点,则应同时满足①、②两式,从而满足③式。即方程③表示的直线过两曲线的交点,又因这样的直线只有一条,故直线③为所求。
This is an exercise in the textbook: Find the linear equation passing through the intersection of the two curves x~2+y~2+3x-y=01 and 3x~2+3y~2+2x+y=02. During the period of enlightenment, the points of intersection can be solved first, and then the linear equation can be obtained: From 1×3-2, 7x-4y=03 can be obtained and 1,2 can be resolved: x_1=0, y_1=0; x_2=-4 /13,y=-7/13. The resulting straight line equation: 7x-4y=0 4 compares 3, 4 and finds that 3 to 4 are loops, and then go back to study formula 3 as the required principle; if (x_1, y_1), (x_2, Y_2) is the intersection point of the two curves, then it should satisfy both the 1 and 2 formulas and satisfy the 3 formula. That is, the straight line represented by Equation 3 crosses the intersection of the two curves, and because there is only one such straight line, the straight line 3 is the desired one.