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

The solution to -((x+2)^2)/4+5y^2=1 is Hyperbola with (h,k)=(-2,0),a=sqrt(1/5),b=2

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

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

(h, k) = (- 2, 0), a = \frac{1}{5}, b = 2

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

\frac{( y - 0 ) ^{2}}{( \frac{1}{5} ) ^{2}} - \frac{( x - ( - 2 ) ) ^{2}}{2 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (- 2, 0), a = \frac{1}{5}, b = 2

9 x^{2} - 4 y^{2} - 54 x - 40 y - 55 = 0

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

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

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

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

What is -((x+2)^2)/4+5y^2=1 ?
The solution to -((x+2)^2)/4+5y^2=1 is Hyperbola with (h,k)=(-2,0),a=sqrt(1/5),b=2