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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

(- \frac{p}{( 2 - q )}) (- \frac{( p - 1 )}{q}) = - 1

2 x^{2} - 2 y^{2} - 20 x + 8 y + 34 = 0

y^{2} - 16 x^{2} + 64 x - 200 = 0

4 x^{2} - 5 y^{2} - 24 y - 16 = 0

- 3 x^{2} + y^{2} - 12 = 0

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