What is 5(y+3)^2-4(x-1)^2=20 ?

The solution to 5(y+3)^2-4(x-1)^2=20 is Hyperbola with (h,k)=(1,-3),a=2,b=sqrt(5)

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5(y+3)^2-4(x-1)^2=20

5 (y + 3)^{2} - 4 (x - 1)^{2} = 20

(h, k) = (1, - 3), a = 2, b = 5

Solution steps

5 (y + 3)^{2} - 4 (x - 1)^{2} = 20

Rewrite

5 (y + 3)^{2} - 4 (x - 1)^{2} = 20

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (1, - 3), a = 2, b = 5

\frac{( x - 3 ) ^{2}}{6} - \frac{( y - 1 ) ^{2}}{12} = 1

\frac{( y - 1 ) ^{2}}{36} - \frac{( x - 2 ) ^{2}}{64} = 1

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

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

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

What is 5(y+3)^2-4(x-1)^2=20 ?
The solution to 5(y+3)^2-4(x-1)^2=20 is Hyperbola with (h,k)=(1,-3),a=2,b=sqrt(5)