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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

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

\frac{x ^{2}}{25} - \frac{y ^{2}}{100} = 1

d^{2} = (3 - 20 t)^{2} + (4 - 30 t)^{2}

x^{2} - y^{2} - 4 x - 6 y - 14 = 0

x^{2} - y^{2} - 6 x + 16 y - 119 = 0

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