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

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

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

foci

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

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

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Compute

c :

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

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

Refine

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

y^{2} - 4 x^{2} + 4 x - 2 y - 4 = 0

x^{2} + y^{2} + 2 x = 0

y^{2} = - 4 x

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

\frac{x ^{2}}{9} - \frac{y ^{2}}{64} = 1

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