What is the center ((x-4)^2}{36}-\frac{(y-2)^2)/9 =1 ?

The center ((x-4)^2}{36}-\frac{(y-2)^2)/9 =1 is (4,2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

center ((x-4)^2}{36}-\frac{(y-2)^2)/9 =1

center

\frac{( x - 4 ) ^{2}}{36} - \frac{( y - 2 ) ^{2}}{9} = 1

(4, 2)

Solution steps

\frac{( x - 4 ) ^{2}}{36} - \frac{( y - 2 ) ^{2}}{9} = 1

Rewrite

\frac{( x - 4 ) ^{2}}{36} - \frac{( y - 2 ) ^{2}}{9} = 1

in the form of a standard hyperbola equation

\frac{( x - 4 ) ^{2}}{6 ^{2}} - \frac{( y - 2 ) ^{2}}{3 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (4, 2), a = 6, b = 3

And the center is:

(4, 2)

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

4 x^{2} - 9 y^{2} + 32 x + 36 y + 64 = 0

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

\frac{( y - 2 ) ^{2}}{9} - \frac{( x - 1 ) ^{2}}{16} = 1

\frac{x ^{2}}{36} - \frac{y ^{2}}{9} = 1

What is the center ((x-4)^2}{36}-\frac{(y-2)^2)/9 =1 ?
The center ((x-4)^2}{36}-\frac{(y-2)^2)/9 =1 is (4,2)