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

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

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

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

(h, k) = (- 2, - 3), a = \frac{5}{2}, b = 3

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (- 2, - 3), a = \frac{5}{2}, b = 3

4 x^{2} - y^{2} - 8 x - 2 y - 13 = 0

\frac{y ^{2}}{2} - \frac{x ^{2}}{\frac{8}{7}} = 1

4 x^{2} - y^{2} - 4 x + 3 y - 26 = 0

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

- 9 x^{2} + 4 y^{2} - 144 = 0

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