What is ((x-3)^2}{25}-\frac{(y+2)^2)/9 =1 ?

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

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((x-3)^2}{25}-\frac{(y+2)^2)/9 =1

\frac{( x - 3 ) ^{2}}{25} - \frac{( y + 2 ) ^{2}}{9} = 1

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

Solution steps

\frac{( x - 3 ) ^{2}}{25} - \frac{( y + 2 ) ^{2}}{9} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{25} - \frac{( y + 2 ) ^{2}}{9} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

x^{2} - y^{2} - 8 x - 8 y - 15 = 0

\frac{a + 10}{b} = \frac{b}{a}

(x - 2)^{2} - (y + 5)^{2} = 18

\frac{4 y ^{2}}{9} - \frac{4 x ^{2}}{27} = 1

2 x^{2} - 3 y^{2} + 8 x + 6 y - 115 = 0

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