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foci
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Calculate Hyperbola properties
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foci ((x+2)^2}{16}-\frac{(y-3)^2)/9 =1foci ((x+2)^2)/(49)-((y-2)^2)/(45)=1foci ((y-3)^2)/(8^2)-((x-2)^2)/(8^2)=1foci (x^2}{18}-\frac{y^2)/7 =1foci ((x^2))/4-0.14y^2=1
Frequently Asked Questions (FAQ)
What is the foci 289y^2-x^2=1 ?
The foci 289y^2-x^2=1 is (0,(sqrt(290))/(17)),(0,-(sqrt(290))/(17))