What is ((x+3)^2)/4-((y-1)^2)/5 =1 ?

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

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((x+3)^2)/4-((y-1)^2)/5 =1

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

(h, k) = (- 3, 1), a = 2, b = 5

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (- 3, 1), a = 2, b = 5

16 y^{(2)} - x^{(2)} + 2 x + 64 y - 1 = 0

(\frac{x}{7})^{2} - (\frac{y}{9})^{2} = 1

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

\frac{( y - 7 ) ^{2}}{36} - \frac{( x - 3 ) ^{2}}{64} = 1

- 3 (x^{2} - 2 x + 1) + (y^{2} + 4 y + 2) - 2 = - 11

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