What is ((x-2)^2)/4-((y+3)^2)/5 =1 ?

The solution to ((x-2)^2)/4-((y+3)^2)/5 =1 is Hyperbola with (h,k)=(2,-3),a=2,b=sqrt(5)

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((x-2)^2)/4-((y+3)^2)/5 =1

\frac{( x - 2 ) ^{2}}{4} - \frac{( y + 3 ) ^{2}}{5} = 1

(h, k) = (2, - 3), a = 2, b = 5

Solution steps

\frac{( x - 2 ) ^{2}}{4} - \frac{( y + 3 ) ^{2}}{5} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{4} - \frac{( y + 3 ) ^{2}}{5} = 1

in the form of a standard hyperbola equation

\frac{( x - 2 ) ^{2}}{2 ^{2}} - \frac{( y - ( - 3 ) ) ^{2}}{( 5 ) ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (2, - 3), a = 2, b = 5

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

x^{2} - y^{2} = 36

x^{2} - 4 y^{2} + 6 x + 8 y = - 1

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

\frac{x ^{2}}{4 + y ^{2}} = 1

What is ((x-2)^2)/4-((y+3)^2)/5 =1 ?
The solution to ((x-2)^2)/4-((y+3)^2)/5 =1 is Hyperbola with (h,k)=(2,-3),a=2,b=sqrt(5)