What is the foci (x^2)/(16)-(y^2)/(20)=1 ?

The foci (x^2)/(16)-(y^2)/(20)=1 is (6,0),(-6,0)

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foci (x^2)/(16)-(y^2)/(20)=1

foci

\frac{x ^{2}}{16} - \frac{y ^{2}}{20} = 1

(6, 0), (- 6, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{16} - \frac{y ^{2}}{20} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 4, b = 25

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

Compute

c :

c = 4^{2} + (25)^{2} : 6

(0 + 6, 0), (0 - 6, 0)

Refine

(6, 0), (- 6, 0)

9 x^{2} - 4 y^{2} - 72 x = 0

f (x) = 2 x^{2} - x - 6

x^{2} - 4 y = 0

y^{2} = 4 x

3 x^{2} + 12 x + 4 y^{2} - 40 y + 100 = 0

What is the foci (x^2)/(16)-(y^2)/(20)=1 ?
The foci (x^2)/(16)-(y^2)/(20)=1 is (6,0),(-6,0)