What is the asymptotes of (y)^2-((x+1)/2)^2=1 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

asymptotes of (y)^2-((x+1)/2)^2=1

asymptotes

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

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

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

(h, k) = (- 1, 0), a = 1, b = 2

y = \frac{1}{2} (x - (- 1)) + 0, y = - \frac{1}{2} (x - (- 1)) + 0

Refine

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

9 y^{2} - 49 x^{2} = 441

4 x^{2} - 9 y^{2} - 16 x + 18 y - 7 = 0

64 x^{2} - 100 y^{2} = 6400

\frac{x ^{2}}{3} - \frac{y ^{2}}{1.5} = 1

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

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