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

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

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

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

(h, k) = (- 1, 3), a = 9, b = 16

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

\frac{( x - ( - 1 ) ) ^{2}}{9 ^{2}} - \frac{( y - 3 ) ^{2}}{1 6 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (- 1, 3), a = 9, b = 16

\frac{x ^{2}}{4} - \frac{y ^{2}}{36} = 1

x^{2} + 2 y + x = 2 y^{2} + 3 y + 2 x + 2

- 4 x^{2} + 24 x + 16 y^{2} - 128 y + 156 = 0

(x + y) (x - y) - 2 = 0

\frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1

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