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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

4 y^{2} - 2 x^{2} - 4 y - 15 = 0

\frac{x ^{2}}{2} - \frac{1}{2} = \frac{y ^{2}}{200} - \frac{1}{200}

- y^{2} + 8 z^{2} = - 8

9 x^{2} - 8 y^{2} = 3

\frac{( x + 8 ) ^{2}}{169} - \frac{( y + 9 ) ^{2}}{25} = 1

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