What is -9x^2+4y^2-108x+32y-296=0 ?

The solution to -9x^2+4y^2-108x+32y-296=0 is Hyperbola with (h,k)=(-6,-4),a=3,b=2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-9x^2+4y^2-108x+32y-296=0

- 9 x^{2} + 4 y^{2} - 108 x + 32 y - 296 = 0

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

Solution steps

- 9 x^{2} + 4 y^{2} - 108 x + 32 y - 296 = 0

Rewrite

- 9 x^{2} + 4 y^{2} - 108 x + 32 y - 296 = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

\frac{( x - 1 ) ^{2}}{9} - \frac{y ^{2}}{5} = 100

x^{2} = 2 y^{2} - 2 \cdot 0.01

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

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

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

What is -9x^2+4y^2-108x+32y-296=0 ?
The solution to -9x^2+4y^2-108x+32y-296=0 is Hyperbola with (h,k)=(-6,-4),a=3,b=2