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

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

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

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

(h, k) = (2, - 1), a = 2, b = 5

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (2, - 1), a = 2, b = 5

11 x^{2} - 25 y^{2} + 22 x + 250 y - 889 = 0

x^{2} + 2 x - 100 y^{2} - 1000 y + 2401 = 0

9 x^{2} - 16 y^{2} + 72 x + 96 y + 144 = 0

\frac{y ^{2}}{16} - \frac{x ^{2}}{144} = 1

25 x^{2} - 16 y^{2} + 250 x + 32 y + 109 = 0

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