What is ((y-3)^2)/(16)-((x+4)^2)/(25)=1 ?

The solution to ((y-3)^2)/(16)-((x+4)^2)/(25)=1 is Hyperbola with (h,k)=(-4,3),a=4,b=5

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((y-3)^2)/(16)-((x+4)^2)/(25)=1

\frac{( y - 3 ) ^{2}}{16} - \frac{( x + 4 ) ^{2}}{25} = 1

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

Solution steps

\frac{( y - 3 ) ^{2}}{16} - \frac{( x + 4 ) ^{2}}{25} = 1

Rewrite

\frac{( y - 3 ) ^{2}}{16} - \frac{( x + 4 ) ^{2}}{25} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

2 x^{2} - 2 y^{2} = 3

x^{2} - 2 y^{2} - x = 1

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

x^{2} - 2 x - 4 y^{2} - 40 y = 103

16 X^{2} - 25 Y^{2} + 96 X + 544 = 0

What is ((y-3)^2)/(16)-((x+4)^2)/(25)=1 ?
The solution to ((y-3)^2)/(16)-((x+4)^2)/(25)=1 is Hyperbola with (h,k)=(-4,3),a=4,b=5