What is y+y^2=x^2+4 ?

The solution to y+y^2=x^2+4 is Hyperbola with (h,k)=(0,-1/2),a=(sqrt(17))/2 ,b=(sqrt(17))/2

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y+y^2=x^2+4

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

(h, k) = (0, - \frac{1}{2}), a = \frac{17}{2}, b = \frac{17}{2}

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

\frac{( y - ( - \frac{1}{2} ) ) ^{2}}{( \frac{17}{2} ) ^{2}} - \frac{( x - 0 ) ^{2}}{( \frac{17}{2} ) ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (0, - \frac{1}{2}), a = \frac{17}{2}, b = \frac{17}{2}

y + y^{2} = x^{2} + 6

x^{2} - y^{2} = x - y + 2018

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

64 Y^{2} - 25 X^{2} = 1600

\frac{( y - 4 ) ^{2}}{25} - \frac{( x + 6 ) ^{2}}{9} = 1

What is y+y^2=x^2+4 ?
The solution to y+y^2=x^2+4 is Hyperbola with (h,k)=(0,-1/2),a=(sqrt(17))/2 ,b=(sqrt(17))/2