What is -x^2+4y^2-16y+12=0 ?

The solution to -x^2+4y^2-16y+12=0 is Hyperbola with (h,k)=(0,2),a=1,b=2

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-x^2+4y^2-16y+12=0

- x^{2} + 4 y^{2} - 16 y + 12 = 0

(h, k) = (0, 2), a = 1, b = 2

Solution steps

- x^{2} + 4 y^{2} - 16 y + 12 = 0

Rewrite

- x^{2} + 4 y^{2} - 16 y + 12 = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (0, 2), a = 1, b = 2

\frac{( x - 3 ) ^{2}}{25} - \frac{( y - 2 ) ^{2}}{40} = 1

\frac{( x - 1 ) ^{2}}{4} - \frac{( y + 2 ) ^{2}}{13} = 1

10 0^{2} + 15 t^{2} = d^{2}

x^{2} - y^{2} - 6 x - 2 y - 10 = 0

y^{2} = x^{2} - 8 x

What is -x^2+4y^2-16y+12=0 ?
The solution to -x^2+4y^2-16y+12=0 is Hyperbola with (h,k)=(0,2),a=1,b=2