What is the vertices-x^2+4y^2+2x+24y+31=0 ?

The vertices-x^2+4y^2+2x+24y+31=0 is (1,-2),(1,-4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices-x^2+4y^2+2x+24y+31=0

vertices

- x^{2} + 4 y^{2} + 2 x + 24 y + 31 = 0

(1, - 2), (1, - 4)

Solution steps

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

Calculate Hyperbola properties

- x^{2} + 4 y^{2} + 2 x + 24 y + 31 = 0 :

Up-down Hyperbola with

(h, k) = (1, - 3), a = 1, b = 2

(1, - 3 + 1), (1, - 3 - 1)

Refine

(1, - 2), (1, - 4)

9 x^{2} - 4 y^{2} = 144

4 y^{2} - 9 x^{2} + 40 y - 540 x - 8036 = 0

11 x^{2} - 25 y^{2} + 22 x + 250 y - 889 = 0

4 x^{2} - 45 y^{2} = 180

\frac{x ^{2}}{16} - \frac{y ^{2}}{48} = 1

What is the vertices-x^2+4y^2+2x+24y+31=0 ?
The vertices-x^2+4y^2+2x+24y+31=0 is (1,-2),(1,-4)