What is the vertices x^2-y^2+2y=0 ?

The vertices x^2-y^2+2y=0 is (0,2),(0,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices x^2-y^2+2y=0

vertices

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

(0, 2), (0, 0)

Solution steps

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

Calculate Hyperbola properties

x^{2} - y^{2} + 2 y = 0 :

Up-down Hyperbola with

(h, k) = (0, 1), a = 1, b = 1

(0, 1 + 1), (0, 1 - 1)

Refine

(0, 2), (0, 0)

\frac{x ^{2}}{64} - \frac{y ^{2}}{36} = 1

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

\frac{y ^{2}}{25} - \frac{x ^{2}}{144} = 1

4 x^{2} - 9 y^{2} - 24 x - 36 y - 36 = 0

9 x^{2} - 25 y^{2} = 225