What is ((y+4)^2)/(64)-((x+3)^2)/(25)=1 ?

The solution to ((y+4)^2)/(64)-((x+3)^2)/(25)=1 is Hyperbola with (h,k)=(-3,-4),a=8,b=5

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((y+4)^2)/(64)-((x+3)^2)/(25)=1

\frac{( y + 4 ) ^{2}}{64} - \frac{( x + 3 ) ^{2}}{25} = 1

(h, k) = (- 3, - 4), a = 8, b = 5

Solution steps

\frac{( y + 4 ) ^{2}}{64} - \frac{( x + 3 ) ^{2}}{25} = 1

Rewrite

\frac{( y + 4 ) ^{2}}{64} - \frac{( x + 3 ) ^{2}}{25} = 1

in the form of a standard hyperbola equation

\frac{( y - ( - 4 ) ) ^{2}}{8 ^{2}} - \frac{( x - ( - 3 ) ) ^{2}}{5 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (- 3, - 4), a = 8, b = 5

\frac{( 4 x ^{2} - 4 y ^{2} - 4 )}{( x ^{2} + y ^{2} + 1 ) ^{2}} = 0

f (x) (f (x)) = 2 x^{2} - 3 x + 2

\frac{y ^{2}}{21} - \frac{x ^{2}}{13} = 1

8 x^{2} - 15 = 4 y^{2} + 1

\frac{x ^{2}}{400} - \frac{y ^{2}}{441} = 1

What is ((y+4)^2)/(64)-((x+3)^2)/(25)=1 ?
The solution to ((y+4)^2)/(64)-((x+3)^2)/(25)=1 is Hyperbola with (h,k)=(-3,-4),a=8,b=5