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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

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

x^{2} - 9 y^{2} + 2 x + 36 y - 44 = 0

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

x^{2} + (6 - 6) y^{2} = 1

15485863 = 4 x^{2} - y^{2}

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