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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

\frac{( x - 2 ) ^{2}}{81} - \frac{y ^{2}}{49} = 1

3 x^{2} - 2 y^{2} + 32 y - 134 = 0

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

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

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

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