What is the vertices (y^2)/4-(x^2)/(25)=1 ?

The vertices (y^2)/4-(x^2)/(25)=1 is (0,2),(0,-2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices (y^2)/4-(x^2)/(25)=1

vertices

\frac{y ^{2}}{4} - \frac{x ^{2}}{25} = 1

(0, 2), (0, - 2)

Solution steps

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

Calculate Hyperbola properties

\frac{y ^{2}}{4} - \frac{x ^{2}}{25} = 1 :

Up-down Hyperbola with

(h, k) = (0, 0), a = 2, b = 5

(0, 0 + 2), (0, 0 - 2)

Refine

(0, 2), (0, - 2)

\frac{x ^{2}}{7} - \frac{7 y ^{2}}{42} = 1

x^{2} = \frac{9}{16} y^{2} + 9

9 x^{2} - 4 y^{2} - 36 x + 8 y - 4 = 0

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

y^{2} - x^{2} = 1

What is the vertices (y^2)/4-(x^2)/(25)=1 ?
The vertices (y^2)/4-(x^2)/(25)=1 is (0,2),(0,-2)