y^2=8+4x^2

Solution

y^{2} = 8 + 4 x^{2}

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

Solution steps

y^{2} = 8 + 4 x^{2}

Rewrite

y^{2} = 8 + 4 x^{2}

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

(x - 2)^{2} - (y - 4)^{2} = 2

x^{2} - y^{2} \leq 16

x^{2} + 25 = y^{2}

x^{2} - y^{2} - 2 x - y + 1 = 0

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

What is y^2=8+4x^2 ?
The solution to y^2=8+4x^2 is Hyperbola with (h,k)=(0,0),a=2sqrt(2),b=sqrt(2)