What is the foci (x^2)/4-(y^2)/(16)=1 ?

The foci (x^2)/4-(y^2)/(16)=1 is (2sqrt(5),0),(-2sqrt(5),0)

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foci (x^2)/4-(y^2)/(16)=1

foci

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

(25, 0), (- 25, 0)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (0, 0), a = 2, b = 4

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

Compute

c :

c = 2^{2} + 4^{2} : 25

(0 + 25, 0), (0 - 25, 0)

Refine

(25, 0), (- 25, 0)

x^{2} - y^{2} = 3

9 x^{2} + 4 y^{2} - 18 x + 16 y - 11 = 0

x^{2} + 16 y^{2} = 16

f (x) = x^{2} - 2 x + 1

y^{2} = - 10 x

What is the foci (x^2)/4-(y^2)/(16)=1 ?
The foci (x^2)/4-(y^2)/(16)=1 is (2sqrt(5),0),(-2sqrt(5),0)