What is ((x-4)^2)/(16)-((y+5)^2)/(25)=1 ?

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

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((x-4)^2)/(16)-((y+5)^2)/(25)=1

\frac{( x - 4 ) ^{2}}{16} - \frac{( y + 5 ) ^{2}}{25} = 1

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

Solution steps

\frac{( x - 4 ) ^{2}}{16} - \frac{( y + 5 ) ^{2}}{25} = 1

Rewrite

\frac{( x - 4 ) ^{2}}{16} - \frac{( y + 5 ) ^{2}}{25} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

4 x^{2} - 3 y^{2} + 5 y - 6 y + 2 x = 7

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

y^{2} - x^{2} = 121

\frac{x ^{2}}{60} - \frac{y ^{2}}{40} = 1

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

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