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Dick
#10
Sep23-10, 10:36 AM
Sci Advisor
HW Helper
Thanks
P: 25,244
It's really not so hard to solve for the intersection points. In fact, it's pretty easy. You've got y=ax/b. Put that into the quadratics. But I think the easiest way is to multiply your two derivatives (a-2x)/(2y) and (2x)/(b-2y) and see if you can show it's (-1) just using the original equations. BTW, one of the intersection points is clearly (0,0). You might need to make a special argument there. Sometimes this sort of question does take some cleverness to see an easy method.