What is ((x-3)^2)/(16)-((y+2)^2)/(16)=1 ?

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

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((x-3)^2)/(16)-((y+2)^2)/(16)=1

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

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

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

x^{2} - y^{2} = \frac{64}{5}

y^{2} = x^{2} - 5 x + 4

a^{2} - b^{2} = 50

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

4 x^{2} - 24 x - 25 y^{2} + 250 y - 489 = 0

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