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foci
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Calculate Hyperbola properties
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foci+(x^2)/4-(y^2)/4 =1foci-4x^2+y^2=36foci (y^2)/4-(x^2)/(21)=1foci x^2-y^2=100foci ((x+2)^2)/4-((y-1)^2)/9 =1
Frequently Asked Questions (FAQ)
What is the foci (y^2)/(64)-(x^2)/(49)=1 ?
The foci (y^2)/(64)-(x^2)/(49)=1 is (0,sqrt(113)),(0,-sqrt(113))