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Frequently Asked Questions (FAQ)
What is the foci ((y-1)^2)/(1/4)-((x+1)^2)/(1/16)=1 ?
- The foci ((y-1)^2)/(1/4)-((x+1)^2)/(1/16)=1 is (-1,(4+sqrt(5))/4),(-1,(4-sqrt(5))/4)