What is (y+2)^2=2x^2+8 ?

The solution to (y+2)^2=2x^2+8 is Hyperbola with (h,k)=(0,-2),a=2sqrt(2),b=2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(y+2)^2=2x^2+8

(y + 2)^{2} = 2 x^{2} + 8

(h, k) = (0, - 2), a = 22, b = 2

Solution steps

(y + 2)^{2} = 2 x^{2} + 8

Rewrite

(y + 2)^{2} - 2 x^{2} - 8 = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (0, - 2), a = 22, b = 2

\frac{( y - 8 ) ^{2}}{25} - \frac{( x - 12 ) ^{2}}{16} = 1

(a - b) (a + b) = 3000

y^{2} = x^{2} + 576

3 4^{2} + x^{2} = y^{2}

y^{(2)} - 9 x^{2} + 8 x - 8 y - 2 = 0

What is (y+2)^2=2x^2+8 ?
The solution to (y+2)^2=2x^2+8 is Hyperbola with (h,k)=(0,-2),a=2sqrt(2),b=2