What is ((y-3)^2)/4-((x+1)^2)/(16)=1 ?

The solution to ((y-3)^2)/4-((x+1)^2)/(16)=1 is Hyperbola with (h,k)=(-1,3),a=2,b=4

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((y-3)^2)/4-((x+1)^2)/(16)=1

\frac{( y - 3 ) ^{2}}{4} - \frac{( x + 1 ) ^{2}}{16} = 1

(h, k) = (- 1, 3), a = 2, b = 4

Solution steps

\frac{( y - 3 ) ^{2}}{4} - \frac{( x + 1 ) ^{2}}{16} = 1

Rewrite

\frac{( y - 3 ) ^{2}}{4} - \frac{( x + 1 ) ^{2}}{16} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (- 1, 3), a = 2, b = 4

\frac{x ^{2}}{4 ^{2}} - \frac{y ^{2}}{3 ^{2}} = 1

x^{2} + 6 = y^{2} + 6 x - 2 y

4 x^{2} - 24 x - y^{2} - 2 y = - 31

9 y^{2} - 4 x^{2} + 32 x - 36 y - 64 = 0

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

What is ((y-3)^2)/4-((x+1)^2)/(16)=1 ?
The solution to ((y-3)^2)/4-((x+1)^2)/(16)=1 is Hyperbola with (h,k)=(-1,3),a=2,b=4