What is ((x-2)^2)/4-((y+2)^2)/5 =1 ?

The solution to ((x-2)^2)/4-((y+2)^2)/5 =1 is Hyperbola with (h,k)=(2,-2),a=2,b=sqrt(5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

((x-2)^2)/4-((y+2)^2)/5 =1

\frac{( x - 2 ) ^{2}}{4} - \frac{( y + 2 ) ^{2}}{5} = 1

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

Solution steps

\frac{( x - 2 ) ^{2}}{4} - \frac{( y + 2 ) ^{2}}{5} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{4} - \frac{( y + 2 ) ^{2}}{5} = 1

in the form of a standard hyperbola equation

\frac{( x - 2 ) ^{2}}{2 ^{2}} - \frac{( y - ( - 2 ) ) ^{2}}{( 5 ) ^{2}} = 1

Therefore Hyperbola properties are:

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

What is ((x-2)^2)/4-((y+2)^2)/5 =1 ?
The solution to ((x-2)^2)/4-((y+2)^2)/5 =1 is Hyperbola with (h,k)=(2,-2),a=2,b=sqrt(5)