What is the foci ((x)^2)/(25/4)-((y)^2)/(11/4)=1 ?

The foci ((x)^2)/(25/4)-((y)^2)/(11/4)=1 is (3,0),(-3,0)

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foci ((x)^2)/(25/4)-((y)^2)/(11/4)=1

foci

(3, 0), (- 3, 0)

Solution steps

(h + c, k), (h - c, k)

Calculate Hyperbola properties

\frac{x ^{2}}{\frac{25}{4}} - \frac{y ^{2}}{\frac{11}{4}} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = \frac{5}{2}, b = \frac{11}{2}

(0 + c, 0), (0 - c, 0)

Compute

c :

c = (\frac{5}{2})^{2} + (\frac{11}{2})^{2} : 3

(0 + 3, 0), (0 - 3, 0)

Refine

(3, 0), (- 3, 0)

What is the foci ((x)^2)/(25/4)-((y)^2)/(11/4)=1 ?
The foci ((x)^2)/(25/4)-((y)^2)/(11/4)=1 is (3,0),(-3,0)