What is ((x+3)^2}{25}-\frac{(y-4)^2)/9 =1 ?

The solution to ((x+3)^2}{25}-\frac{(y-4)^2)/9 =1 is Hyperbola with (h,k)=(-3,4),a=5,b=3

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((x+3)^2}{25}-\frac{(y-4)^2)/9 =1

\frac{( x + 3 ) ^{2}}{25} - \frac{( y - 4 ) ^{2}}{9} = 1

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

Solution steps

\frac{( x + 3 ) ^{2}}{25} - \frac{( y - 4 ) ^{2}}{9} = 1

Rewrite

\frac{( x + 3 ) ^{2}}{25} - \frac{( y - 4 ) ^{2}}{9} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

16 x^{2} - 96 x + 144 + 25 y^{2} = 400

x^{2} = 20 y

x^{2} = 4 y

x^{2} + y^{2} = 8

x^{2} - 6 x + y^{2} - 32 y = 0

What is ((x+3)^2}{25}-\frac{(y-4)^2)/9 =1 ?
The solution to ((x+3)^2}{25}-\frac{(y-4)^2)/9 =1 is Hyperbola with (h,k)=(-3,4),a=5,b=3