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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

y^{2} = \frac{4 x ^{2}}{9} - 9

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

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

f (x) f (x) = \frac{15 x ^{3} - 3 x ^{2} - 12 x}{3 x}

\frac{y ^{2}}{100} - \frac{x ^{2}}{49} = 1

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