What is the foci (x^2)/(25)-(y^2)/(16)=1 ?

The foci (x^2)/(25)-(y^2)/(16)=1 is (sqrt(41),0),(-sqrt(41),0)

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foci (x^2)/(25)-(y^2)/(16)=1

foci

\frac{x ^{2}}{25} - \frac{y ^{2}}{16} = 1

(41, 0), (- 41, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{25} - \frac{y ^{2}}{16} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 5, b = 4

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

Compute

c :

c = 5^{2} + 4^{2} : 41

(0 + 41, 0), (0 - 41, 0)

Refine

(41, 0), (- 41, 0)

f (x) = x^{2} - 4 x - 5

\frac{y ^{2}}{16} - \frac{x ^{2}}{4} = 1

x^{2} - 2 x + 8 y + 9 = 0

x^{2} - 5 y = 0

\frac{x ^{2}}{9} + \frac{y ^{2}}{16} = 1

What is the foci (x^2)/(25)-(y^2)/(16)=1 ?
The foci (x^2)/(25)-(y^2)/(16)=1 is (sqrt(41),0),(-sqrt(41),0)