What is ((x+3)^2)/9-((y-3)^2)/(16)=1 ?

The solution to ((x+3)^2)/9-((y-3)^2)/(16)=1 is Hyperbola with (h,k)=(-3,3),a=3,b=4

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((x+3)^2)/9-((y-3)^2)/(16)=1

\frac{( x + 3 ) ^{2}}{9} - \frac{( y - 3 ) ^{2}}{16} = 1

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

Solution steps

\frac{( x + 3 ) ^{2}}{9} - \frac{( y - 3 ) ^{2}}{16} = 1

Rewrite

\frac{( x + 3 ) ^{2}}{9} - \frac{( y - 3 ) ^{2}}{16} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

4 x^{2} + 8 x - 4 y^{2} + 12 y - 3 = 0

\frac{( x + 4 ) ^{2}}{16} - \frac{( y - 2 ) ^{2}}{36} = 1

12 y^{2} + 72 y - 4 x^{2} + 16 x + 44 = 0

(x - 1)^{2} - (y - 4)^{2} = 16

y^{2} = x^{2} + 289

What is ((x+3)^2)/9-((y-3)^2)/(16)=1 ?
The solution to ((x+3)^2)/9-((y-3)^2)/(16)=1 is Hyperbola with (h,k)=(-3,3),a=3,b=4