What is the center 121-32x-8x^2-90y+9y^2=0 ?

The center 121-32x-8x^2-90y+9y^2=0 is (-2,5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

center 121-32x-8x^2-90y+9y^2=0

center

121 - 32 x - 8 x^{2} - 90 y + 9 y^{2} = 0

(- 2, 5)

Solution steps

121 - 32 x - 8 x^{2} - 90 y + 9 y^{2} = 0

Rewrite

121 - 32 x - 8 x^{2} - 90 y + 9 y^{2} = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (- 2, 5), a = 22, b = 3

And the center is:

(- 2, 5)

y^{2} - 4 x^{2} = 36

10 x^{2} - 6 y^{2} - 20 x - 24 y + 46 = 0

- 4 x^{2} + 24 x + 16 y^{2} - 128 y + 156 = 0

\frac{( x - 1 ) ^{2}}{9} - \frac{( y + 6 ) ^{2}}{4} = 1

25 x^{2} - 4 y^{2} - 100 x - 16 y - 16 = 0

What is the center 121-32x-8x^2-90y+9y^2=0 ?
The center 121-32x-8x^2-90y+9y^2=0 is (-2,5)