What is ((x-2)^2)/9-((y+1)^2)/(16)=1 ?

The solution to ((x-2)^2)/9-((y+1)^2)/(16)=1 is Hyperbola with (h,k)=(2,-1),a=3,b=4

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((x-2)^2)/9-((y+1)^2)/(16)=1

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

(h, k) = (2, - 1), a = 3, b = 4

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (2, - 1), a = 3, b = 4

x^{2} - y^{2} = 96

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

3 y^{2} - x^{2} + 24 y - 2 x = 0

x^{2} - y^{2} = 60

x^{2} - y^{2} = 67

What is ((x-2)^2)/9-((y+1)^2)/(16)=1 ?
The solution to ((x-2)^2)/9-((y+1)^2)/(16)=1 is Hyperbola with (h,k)=(2,-1),a=3,b=4