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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

f (x) = x^{2} - 2 x + 4

x^{2} - 24 x - 12

16 x^{2} + 9 y^{2} = 144

- x^{2} + y^{2} = 1

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

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