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

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

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

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

(h, k) = (- 4, 2), a = 5, b = 2

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (- 4, 2), a = 5, b = 2

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

9 x^{2} - 16 = (4 x + s) (4 x - s)

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

y^{2} = (x - 4)^{2} + 4

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

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