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

9 x^{2} - 16 y^{2} + 36 x + 96 y - 252 = 0

x^{2} - 6 x - 2 y^{2} + 7 = 0

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

(\frac{y}{100})^{2} - \frac{x ^{2}}{49} = 1

(x + y) (x - y) = 60.49

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