What is the foci (x^2)/(25)-(y^2)/(121)=1 ?

The foci (x^2)/(25)-(y^2)/(121)=1 is (sqrt(146),0),(-sqrt(146),0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci (x^2)/(25)-(y^2)/(121)=1

foci

\frac{x ^{2}}{25} - \frac{y ^{2}}{121} = 1

(146, 0), (- 146, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{25} - \frac{y ^{2}}{121} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 5, b = 11

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

Compute

c :

c = 5^{2} + 1 1^{2} : 146

(0 + 146, 0), (0 - 146, 0)

Refine

(146, 0), (- 146, 0)

5 y^{2} - 4 x^{2} - 80 = 0

y^{2} + 2 y = 4 x^{2} + 3

9 x^{2} - 4 y^{2} - 18 x + 16 y - 43 = 0

\frac{x ^{2}}{36} - \frac{y ^{2}}{81} = 1

\frac{x ^{2}}{100} - \frac{y ^{2}}{36} = 1

What is the foci (x^2)/(25)-(y^2)/(121)=1 ?
The foci (x^2)/(25)-(y^2)/(121)=1 is (sqrt(146),0),(-sqrt(146),0)