What is ((x+2)^2)/4-((y-2)^2)/(25)=1 ?

The solution to ((x+2)^2)/4-((y-2)^2)/(25)=1 is Hyperbola with (h,k)=(-2,2),a=2,b=5

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((x+2)^2)/4-((y-2)^2)/(25)=1

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

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

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

2 x^{2} - 3 y^{2} - 12 x - 24 y + 60 = 0

9 x^{2} - 16 y^{2} - 36 = 0

3 x^{2} - y^{2} - 12 = 0

\frac{y ^{2}}{25} - \frac{x ^{2}}{25} = 1

2 x^{(2)} - 2 y^{(2)} - 20 x + 16 y + 16 = 0

What is ((x+2)^2)/4-((y-2)^2)/(25)=1 ?
The solution to ((x+2)^2)/4-((y-2)^2)/(25)=1 is Hyperbola with (h,k)=(-2,2),a=2,b=5