What is ((y-3)^2)/(25)-((x+4)^2)/(36)=1 ?

The solution to ((y-3)^2)/(25)-((x+4)^2)/(36)=1 is Hyperbola with (h,k)=(-4,3),a=5,b=6

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((y-3)^2)/(25)-((x+4)^2)/(36)=1

\frac{( y - 3 ) ^{2}}{25} - \frac{( x + 4 ) ^{2}}{36} = 1

(h, k) = (- 4, 3), a = 5, b = 6

Solution steps

\frac{( y - 3 ) ^{2}}{25} - \frac{( x + 4 ) ^{2}}{36} = 1

Rewrite

\frac{( y - 3 ) ^{2}}{25} - \frac{( x + 4 ) ^{2}}{36} = 1

in the form of a standard hyperbola equation

\frac{( y - 3 ) ^{2}}{5 ^{2}} - \frac{( x - ( - 4 ) ) ^{2}}{6 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (- 4, 3), a = 5, b = 6

- 4 x^{2} + y^{2} = 64

y^{2} - 25 x^{2} + 4 y + 50 x - 46 = 0

\frac{y ^{2}}{20} - \frac{x ^{2}}{1} = 1

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

9 x^{2} - 5 y^{2} + 5 = 0

What is ((y-3)^2)/(25)-((x+4)^2)/(36)=1 ?
The solution to ((y-3)^2)/(25)-((x+4)^2)/(36)=1 is Hyperbola with (h,k)=(-4,3),a=5,b=6