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

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

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y^2<= x^2+y

y^{2} \leq x^{2} + y

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

Solution steps

y^{2} \leq x^{2} + y

Rewrite

y^{2} \leq x^{2} + y

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

36 y^{2} = 4 x^{2} + 144

y^{2} - 9 = x^{2}

3 y^{2} - 12 x^{2} - 36 y - 48 x + 48 = 0

\frac{( y + 7 ) ^{2}}{4} - \frac{( x + 2 ) ^{2}}{16} = 1

\frac{( x - 5 ) ^{2}}{4} - \frac{( y + 2 ) ^{2}}{9} = 1

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