What is the center 4x^2-3y^2-8x-8=0 ?

The center 4x^2-3y^2-8x-8=0 is (1,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

center 4x^2-3y^2-8x-8=0

center

4 x^{2} - 3 y^{2} - 8 x - 8 = 0

(1, 0)

Solution steps

4 x^{2} - 3 y^{2} - 8 x - 8 = 0

Rewrite

4 x^{2} - 3 y^{2} - 8 x - 8 = 0

in the form of a standard hyperbola equation

\frac{( x - 1 ) ^{2}}{( 3 ) ^{2}} - \frac{( y - 0 ) ^{2}}{2 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (1, 0), a = 3, b = 2

And the center is:

(1, 0)

0 = - \frac{400}{9} x^{2} + 16 y^{2} + 400

x^{2} - y^{2} + 7 x + 8 y = 5

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

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

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