What is ((y-3)^2}{16}-\frac{(x+4)^2)/9 =1 ?

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

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((y-3)^2}{16}-\frac{(x+4)^2)/9 =1

\frac{( y - 3 ) ^{2}}{16} - \frac{( x + 4 ) ^{2}}{9} = 1

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

Solution steps

\frac{( y - 3 ) ^{2}}{16} - \frac{( x + 4 ) ^{2}}{9} = 1

Rewrite

\frac{( y - 3 ) ^{2}}{16} - \frac{( x + 4 ) ^{2}}{9} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

2 x^{2} - 2 y = 2 y^{2} + 2 y + 2 x - 1

(\frac{x - 8}{3})^{2} - (\frac{y + 3}{4})^{2} = 1

\frac{y ^{2}}{20} - x^{2} = 1

9 x^{2} - 4 y^{2} - 36 x + 72 = 0

\frac{z ^{2}}{4} - y^{2} = 1

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