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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

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

- x^{2} + 4 y^{2} + 2 x + 24 y + 31 = 0

9 x^{2} - 16 y^{2} - 18 x + 64 y - 199 = 0

3631 + 896 x - 54 y = 9 y^{2} - 64 x^{2}

4 x^{2} - 16 y^{2} - 64 = 0

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