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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

137 + 6 y = y^{2} - 24 x - x^{2}

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

4 x^{2} - y^{2} - 12 y + 36 = 0

9 x^{2} - 72 x - 4 y^{2} - 40 y + 44 = 36

4 x^{2} + 24 x - 2 y^{2} + 36 = 36

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