What is the asymptotes of x^2-4y^2+2x+8y-7=0 ?

The asymptotes of x^2-4y^2+2x+8y-7=0 is y= 1/2 (x+1)+1,\quad y=-1/2 (x+1)+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

asymptotes of x^2-4y^2+2x+8y-7=0

asymptotes

x^{2} - 4 y^{2} + 2 x + 8 y - 7 = 0

y = \frac{1}{2} (x + 1) + 1, y = - \frac{1}{2} (x + 1) + 1

Solution steps

y = \pm \frac{b}{a} (x - h) + k

Calculate Hyperbola properties

x^{2} - 4 y^{2} + 2 x + 8 y - 7 = 0 :

Right-left Hyperbola with

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

y = \frac{1}{2} (x - (- 1)) + 1, y = - \frac{1}{2} (x - (- 1)) + 1

Refine

y = \frac{1}{2} (x + 1) + 1, y = - \frac{1}{2} (x + 1) + 1

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

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

y^{2} - x^{2} = 8

x^{2} - 9 y^{2} = 36

(x - 4)^{2} - (y - 5)^{2} = 15

What is the asymptotes of x^2-4y^2+2x+8y-7=0 ?
The asymptotes of x^2-4y^2+2x+8y-7=0 is y= 1/2 (x+1)+1,\quad y=-1/2 (x+1)+1