What is the foci (x^2)/9-(y^2)/(16)=1 ?

The foci (x^2)/9-(y^2)/(16)=1 is (5,0),(-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)/9-(y^2)/(16)=1

foci

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

(5, 0), (- 5, 0)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (0, 0), a = 3, b = 4

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

Compute

c :

c = 3^{2} + 4^{2} : 5

(0 + 5, 0), (0 - 5, 0)

Refine

(5, 0), (- 5, 0)

- 2 y^{2} + x - 4 y + 1 = 0

\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1

x + y^{2} = 4

y^{2} = - 28 x

\frac{x ^{2}}{25} - \frac{y ^{2}}{9} = 1

What is the foci (x^2)/9-(y^2)/(16)=1 ?
The foci (x^2)/9-(y^2)/(16)=1 is (5,0),(-5,0)