What is ((y-5)^2}{16}-\frac{(x+4)^2)/4 =1 ?

The solution to ((y-5)^2}{16}-\frac{(x+4)^2)/4 =1 is Hyperbola with (h,k)=(-4,5),a=4,b=2

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((y-5)^2}{16}-\frac{(x+4)^2)/4 =1

\frac{( y - 5 ) ^{2}}{16} - \frac{( x + 4 ) ^{2}}{4} = 1

(h, k) = (- 4, 5), a = 4, b = 2

Solution steps

\frac{( y - 5 ) ^{2}}{16} - \frac{( x + 4 ) ^{2}}{4} = 1

Rewrite

\frac{( y - 5 ) ^{2}}{16} - \frac{( x + 4 ) ^{2}}{4} = 1

in the form of a standard hyperbola equation

\frac{( y - 5 ) ^{2}}{4 ^{2}} - \frac{( x - ( - 4 ) ) ^{2}}{2 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (- 4, 5), a = 4, b = 2

- 4 x^{2} + 56 x + y^{2} - 188 = 0

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

4 x^{2} - 9 y^{2} + 144 = 0

- 4 x^{2} + 5 y^{2} + 21 x - 5 y + 28 = 0

4 (x - 4)^{2} - 25 (y - 1)^{2} = 100

What is ((y-5)^2}{16}-\frac{(x+4)^2)/4 =1 ?
The solution to ((y-5)^2}{16}-\frac{(x+4)^2)/4 =1 is Hyperbola with (h,k)=(-4,5),a=4,b=2