What is ((x+2)^2)/(16)-((y-3)^2)/(25)=1 ?

The solution to ((x+2)^2)/(16)-((y-3)^2)/(25)=1 is Hyperbola with (h,k)=(-2,3),a=4,b=5

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((x+2)^2)/(16)-((y-3)^2)/(25)=1

\frac{( x + 2 ) ^{2}}{16} - \frac{( y - 3 ) ^{2}}{25} = 1

(h, k) = (- 2, 3), a = 4, b = 5

Solution steps

\frac{( x + 2 ) ^{2}}{16} - \frac{( y - 3 ) ^{2}}{25} = 1

Rewrite

\frac{( x + 2 ) ^{2}}{16} - \frac{( y - 3 ) ^{2}}{25} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (- 2, 3), a = 4, b = 5

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

3 x^{2} - y^{2} + 18 x + 8 y + 38 = 0

0 = 1 + y^{2} - x^{2}

y^{2} = x^{2} + 12 x + 32

- \frac{( x - 2 ) ^{2}}{9} + \frac{( y + 5 ) ^{2}}{7} = 1

What is ((x+2)^2)/(16)-((y-3)^2)/(25)=1 ?
The solution to ((x+2)^2)/(16)-((y-3)^2)/(25)=1 is Hyperbola with (h,k)=(-2,3),a=4,b=5