What is (-(x-5)^2}{16}+\frac{(y+7)^2)/9 =1 ?

The solution to (-(x-5)^2}{16}+\frac{(y+7)^2)/9 =1 is Hyperbola with (h,k)=(5,-7),a=3,b=4

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(-(x-5)^2}{16}+\frac{(y+7)^2)/9 =1

\frac{- ( x - 5 ) ^{2}}{16} + \frac{( y + 7 ) ^{2}}{9} = 1

(h, k) = (5, - 7), a = 3, b = 4

Solution steps

\frac{- ( x - 5 ) ^{2}}{16} + \frac{( y + 7 ) ^{2}}{9} = 1

Rewrite

\frac{- ( x - 5 ) ^{2}}{16} + \frac{( y + 7 ) ^{2}}{9} = 1

in the form of a standard hyperbola equation

\frac{( y - ( - 7 ) ) ^{2}}{3 ^{2}} - \frac{( x - 5 ) ^{2}}{4 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (5, - 7), a = 3, b = 4

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

y^{2} + C = (\frac{3 x ^{2}}{2} - 6 x + C)

16 x^{2} - 25 y^{2} - 160 x = 0

x^{2} - 4 y^{2} - 36 = 0

y^{2} - 9 x^{2} - 6 y + 54 x = 153

What is (-(x-5)^2}{16}+\frac{(y+7)^2)/9 =1 ?
The solution to (-(x-5)^2}{16}+\frac{(y+7)^2)/9 =1 is Hyperbola with (h,k)=(5,-7),a=3,b=4