What is ((x-5)^2)/(25)-((y+2)^2)/(16)=1 ?

The solution to ((x-5)^2)/(25)-((y+2)^2)/(16)=1 is Hyperbola with (h,k)=(5,-2),a=5,b=4

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((x-5)^2)/(25)-((y+2)^2)/(16)=1

\frac{( x - 5 ) ^{2}}{25} - \frac{( y + 2 ) ^{2}}{16} = 1

(h, k) = (5, - 2), a = 5, b = 4

Solution steps

\frac{( x - 5 ) ^{2}}{25} - \frac{( y + 2 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 5 ) ^{2}}{25} - \frac{( y + 2 ) ^{2}}{16} = 1

in the form of a standard hyperbola equation

\frac{( x - 5 ) ^{2}}{5 ^{2}} - \frac{( y - ( - 2 ) ) ^{2}}{4 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (5, - 2), a = 5, b = 4

2 x^{2} - 2 y^{2} = 12

\frac{x ^{2}}{9} - \frac{y ^{2}}{14} = 1

\frac{( y + 7 ) ^{2}}{9} - \frac{( x - 5 ) ^{2}}{16} = 1

\frac{x ^{2}}{16} - \frac{y ^{2}}{12} = 1

3 x^{2} - 2 y^{2} - 12 x + 12 y + 29 = 0

What is ((x-5)^2)/(25)-((y+2)^2)/(16)=1 ?
The solution to ((x-5)^2)/(25)-((y+2)^2)/(16)=1 is Hyperbola with (h,k)=(5,-2),a=5,b=4