What is ((x-2)^2)/7-((y+1)^2)/9 =1 ?

The solution to ((x-2)^2)/7-((y+1)^2)/9 =1 is Hyperbola with (h,k)=(2,-1),a=sqrt(7),b=3

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((x-2)^2)/7-((y+1)^2)/9 =1

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

(h, k) = (2, - 1), a = 7, b = 3

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

\frac{( x - 2 ) ^{2}}{( 7 ) ^{2}} - \frac{( y - ( - 1 ) ) ^{2}}{3 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (2, - 1), a = 7, b = 3

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

8 y^{2} - x^{2} - 16 y + 8 x - 32 = 0

2 x^{2} - y^{2} + 4 x + 2 y - 3 = 0

5 x^{2} - 2 y^{2} = 9

- x^{2} + 2 y^{2} + 28 y + 14 = 0

What is ((x-2)^2)/7-((y+1)^2)/9 =1 ?
The solution to ((x-2)^2)/7-((y+1)^2)/9 =1 is Hyperbola with (h,k)=(2,-1),a=sqrt(7),b=3