What is ((x-3)^2)/(49)-((y+2)^2)/(16)=1 ?

The solution to ((x-3)^2)/(49)-((y+2)^2)/(16)=1 is Hyperbola with (h,k)=(3,-2),a=7,b=4

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((x-3)^2)/(49)-((y+2)^2)/(16)=1

\frac{( x - 3 ) ^{2}}{49} - \frac{( y + 2 ) ^{2}}{16} = 1

(h, k) = (3, - 2), a = 7, b = 4

Solution steps

\frac{( x - 3 ) ^{2}}{49} - \frac{( y + 2 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{49} - \frac{( y + 2 ) ^{2}}{16} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (3, - 2), a = 7, b = 4

3 x^{2} - y^{2} - 2 x + 39 = 0

3 x^{2} - 4 y^{2} + y = 9

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

y^{2} - x^{2} - 4 y + 4 x - 1 = 0

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

What is ((x-3)^2)/(49)-((y+2)^2)/(16)=1 ?
The solution to ((x-3)^2)/(49)-((y+2)^2)/(16)=1 is Hyperbola with (h,k)=(3,-2),a=7,b=4