What is ((x-2)^2)/9-((y+3)^2)/(27)=1 ?

The solution to ((x-2)^2)/9-((y+3)^2)/(27)=1 is Hyperbola with (h,k)=(2,-3),a=3,b=3sqrt(3)

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((x-2)^2)/9-((y+3)^2)/(27)=1

\frac{( x - 2 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{27} = 1

(h, k) = (2, - 3), a = 3, b = 33

Solution steps

\frac{( x - 2 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{27} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{27} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (2, - 3), a = 3, b = 33

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

(\frac{y - 4}{16})^{2} - (\frac{x + 3}{36})^{2} = 1

\frac{( y - 3 ) ^{2}}{25} - \frac{( x + 4 ) ^{2}}{36} = 1

- 4 x^{2} + y^{2} = 64

y^{2} - 25 x^{2} + 4 y + 50 x - 46 = 0

What is ((x-2)^2)/9-((y+3)^2)/(27)=1 ?
The solution to ((x-2)^2)/9-((y+3)^2)/(27)=1 is Hyperbola with (h,k)=(2,-3),a=3,b=3sqrt(3)