What is 4x^2-y^2+8y-32=0 ?

The solution to 4x^2-y^2+8y-32=0 is Hyperbola with (h,k)=(0,4),a=2,b=4

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4x^2-y^2+8y-32=0

4 x^{2} - y^{2} + 8 y - 32 = 0

(h, k) = (0, 4), a = 2, b = 4

Solution steps

4 x^{2} - y^{2} + 8 y - 32 = 0

Rewrite

4 x^{2} - y^{2} + 8 y - 32 = 0

in the form of a standard hyperbola equation

\frac{( x - 0 ) ^{2}}{2 ^{2}} - \frac{( y - 4 ) ^{2}}{4 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (0, 4), a = 2, b = 4

4 x^{2} - 4 y^{2} - 32 x + 16 y + 39 = 0

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

p^{2} + 2021 = q^{2}

\frac{( y - 1 ) ^{2}}{4 ^{2}} - \frac{( x + 3 ) ^{2}}{5 ^{2}} = 1

x^{2} - 9 y^{2} + 4 x + 18 y - 149 = 3

What is 4x^2-y^2+8y-32=0 ?
The solution to 4x^2-y^2+8y-32=0 is Hyperbola with (h,k)=(0,4),a=2,b=4