What is the center ((x-3)^2)/(8^2)-((y-2)^2)/(8^2)=1 ?

The center ((x-3)^2)/(8^2)-((y-2)^2)/(8^2)=1 is (3,2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

center ((x-3)^2)/(8^2)-((y-2)^2)/(8^2)=1

center

\frac{( x - 3 ) ^{2}}{8 ^{2}} - \frac{( y - 2 ) ^{2}}{8 ^{2}} = 1

(3, 2)

Solution steps

\frac{( x - 3 ) ^{2}}{8 ^{2}} - \frac{( y - 2 ) ^{2}}{8 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (3, 2), a = 8, b = 8

And the center is:

(3, 2)

\frac{( x + 2 ) ^{2}}{36} - \frac{( y - 3 ) ^{2}}{16} = 1

x^{2} - y^{2} - 6 x = 7

4 y^{2} - x^{2} = 900

4 x^{2} - 9 y^{2} - 16 x + 18 y - 11 = 0

\frac{( y - 3 ) ^{2}}{11} - \frac{( x + 1 ) ^{2}}{5} = 1

What is the center ((x-3)^2)/(8^2)-((y-2)^2)/(8^2)=1 ?
The center ((x-3)^2)/(8^2)-((y-2)^2)/(8^2)=1 is (3,2)