What is ((y-1)^2}{25}-\frac{(x+2)^2)/9 =1 ?

The solution to ((y-1)^2}{25}-\frac{(x+2)^2)/9 =1 is Hyperbola with (h,k)=(-2,1),a=5,b=3

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((y-1)^2}{25}-\frac{(x+2)^2)/9 =1

\frac{( y - 1 ) ^{2}}{25} - \frac{( x + 2 ) ^{2}}{9} = 1

(h, k) = (- 2, 1), a = 5, b = 3

Solution steps

\frac{( y - 1 ) ^{2}}{25} - \frac{( x + 2 ) ^{2}}{9} = 1

Rewrite

\frac{( y - 1 ) ^{2}}{25} - \frac{( x + 2 ) ^{2}}{9} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (- 2, 1), a = 5, b = 3

25 x^{2} - 4 y^{2} - 100 x - 16 y - 16 = 0

x^{2} - 5 y^{2} = 25

\frac{y ^{2}}{100} - \frac{x ^{2}}{36} = 1

\frac{y ^{2}}{36} - \frac{x ^{2}}{9} = 1

12 x^{2} - 4 y^{2} - 72 x - 16 y + 44 = 0

What is ((y-1)^2}{25}-\frac{(x+2)^2)/9 =1 ?
The solution to ((y-1)^2}{25}-\frac{(x+2)^2)/9 =1 is Hyperbola with (h,k)=(-2,1),a=5,b=3