What is ((x+3)^2)/9-((y-4)^2)/(16)=1 ?

The solution to ((x+3)^2)/9-((y-4)^2)/(16)=1 is Hyperbola with (h,k)=(-3,4),a=3,b=4

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((x+3)^2)/9-((y-4)^2)/(16)=1

\frac{( x + 3 ) ^{2}}{9} - \frac{( y - 4 ) ^{2}}{16} = 1

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

Solution steps

\frac{( x + 3 ) ^{2}}{9} - \frac{( y - 4 ) ^{2}}{16} = 1

Rewrite

\frac{( x + 3 ) ^{2}}{9} - \frac{( y - 4 ) ^{2}}{16} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

y^{2} = \frac{16 - x ^{2}}{- 4}

(\frac{x + 1}{36})^{2} - (\frac{y - 4}{64})^{2} = 1

\frac{( x - 3 ) ^{2}}{36} - \frac{( y - 2 ) ^{2}}{64} = 1

4 x^{2} - 5 y^{2} - 36 y - 36 = 0

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

What is ((x+3)^2)/9-((y-4)^2)/(16)=1 ?
The solution to ((x+3)^2)/9-((y-4)^2)/(16)=1 is Hyperbola with (h,k)=(-3,4),a=3,b=4