What is the foci (x^2)/(16)-(y^2)/(64)=1 ?

The foci (x^2)/(16)-(y^2)/(64)=1 is (4sqrt(5),0),(-4sqrt(5),0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci (x^2)/(16)-(y^2)/(64)=1

foci

\frac{x ^{2}}{16} - \frac{y ^{2}}{64} = 1

(45, 0), (- 45, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{16} - \frac{y ^{2}}{64} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 4, b = 8

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

Compute

c :

c = 4^{2} + 8^{2} : 45

(0 + 45, 0), (0 - 45, 0)

Refine

(45, 0), (- 45, 0)

\frac{( x - 2 ) ^{2}}{9} - \frac{( y + 1 ) ^{2}}{5} = 1

x^{2} - y^{2} \geq 1

\frac{( x - 6 ) ^{2}}{16} - \frac{( y - 2 ) ^{2}}{9} = 1

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

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

What is the foci (x^2)/(16)-(y^2)/(64)=1 ?
The foci (x^2)/(16)-(y^2)/(64)=1 is (4sqrt(5),0),(-4sqrt(5),0)