What is the foci (x^2)/9-(y^2)/(40)=1 ?

The foci (x^2)/9-(y^2)/(40)=1 is (7,0),(-7,0)

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foci (x^2)/9-(y^2)/(40)=1

foci

\frac{x ^{2}}{9} - \frac{y ^{2}}{40} = 1

(7, 0), (- 7, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{9} - \frac{y ^{2}}{40} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 3, b = 210

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

Compute

c :

c = 3^{2} + (210)^{2} : 7

(0 + 7, 0), (0 - 7, 0)

Refine

(7, 0), (- 7, 0)

16 y^{2} - 32 y - 9 x^{2} - 18 x = 137

\frac{x ^{2}}{9} - \frac{y ^{2}}{1} = 1

25 y^{2} - x^{2} = 100

- 3 x^{2} + 2 y^{2} + 6 x - 7 y = 7

\frac{x ^{2}}{36} - \frac{y ^{2}}{16} = 1

What is the foci (x^2)/9-(y^2)/(40)=1 ?
The foci (x^2)/9-(y^2)/(40)=1 is (7,0),(-7,0)