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The third celestial circle and quadratic function problem
Firstly, three known points are directly substituted into the coordinates to solve the conic curve.

0=36a+6b+c 0=4a+2b+c 2 times the root number 3=C, so a=6 times the root number 3, b= minus 3 times the root number 3, and c=2 times the root number 3.

I don't know if I have finished studying in Grade Three, but it seems to be the content of Grade One, but I will tell you the answer first. The center of the circle should be (4,2 times the root number 3).

Then, because the PAB coordinates are all known, there are three possibilities for the coordinates of point D, (8,2 times the root of 3) (0,2 times the root of 3) (4, the root of minus 2 times 3), which are respectively substituted into the equation and do not match, so there is no one.