What is the center ((x-3)^2)/(a^2)-((y-4)^2)/(b^2)=1 ?

The center ((x-3)^2)/(a^2)-((y-4)^2)/(b^2)=1 is (3,4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

center ((x-3)^2)/(a^2)-((y-4)^2)/(b^2)=1

center

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

(3, 4)

Solution steps

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

Therefore Hyperbola properties are:

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

And the center is:

(3, 4)

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

\frac{( y - 2 ) ^{2}}{9} - \frac{( x + 1 ) ^{2}}{25} = 1

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

- 4 x^{2} + y^{2} + 8 y - 20 = 0

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

What is the center ((x-3)^2)/(a^2)-((y-4)^2)/(b^2)=1 ?
The center ((x-3)^2)/(a^2)-((y-4)^2)/(b^2)=1 is (3,4)