What is ((x-3)^2)/4-(y^2)/(25)=1 ?

The solution to ((x-3)^2)/4-(y^2)/(25)=1 is Hyperbola with (h,k)=(3,0),a=2,b=5

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((x-3)^2)/4-(y^2)/(25)=1

\frac{( x - 3 ) ^{2}}{4} - \frac{y ^{2}}{25} = 1

(h, k) = (3, 0), a = 2, b = 5

Solution steps

\frac{( x - 3 ) ^{2}}{4} - \frac{y ^{2}}{25} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{4} - \frac{y ^{2}}{25} = 1

in the form of a standard hyperbola equation

\frac{( x - 3 ) ^{2}}{2 ^{2}} - \frac{( y - 0 ) ^{2}}{5 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (3, 0), a = 2, b = 5

x^{2} - 4 y^{2} - 4 > 0

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

x^{2} = y^{2} + 13

y^{2} - x^{2} = \frac{9}{25}

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

What is ((x-3)^2)/4-(y^2)/(25)=1 ?
The solution to ((x-3)^2)/4-(y^2)/(25)=1 is Hyperbola with (h,k)=(3,0),a=2,b=5