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

The vertices (y^2)/(25)-(x^2)/(16)=1 is (0,5),(0,-5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

vertices

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

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

Solution steps

(h, k + a), (h, k - a)

Calculate Hyperbola properties

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

Up-down Hyperbola with

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

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

Refine

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

\frac{x ^{2}}{169} + \frac{y ^{2}}{25} = 1

2 x - y^{2} = 4

f (x) = x^{2} - 24 x - 12

x^{2} = - 12 y

f (x) = x^{2} + 6 x + 8

What is the vertices (y^2)/(25)-(x^2)/(16)=1 ?
The vertices (y^2)/(25)-(x^2)/(16)=1 is (0,5),(0,-5)