What is 4x^2-8y^2<= 12 ?

The solution to 4x^2-8y^2<= 12 is Hyperbola with (h,k)=(0,0),a=sqrt(3),b=sqrt(3/2)

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4x^2-8y^2<= 12

4 x^{2} - 8 y^{2} \leq 12

(h, k) = (0, 0), a = 3, b = \frac{3}{2}

Solution steps

4 x^{2} - 8 y^{2} \leq 12

Rewrite

4 x^{2} - 8 y^{2} \leq 12

in the form of a standard hyperbola equation

\frac{( x - 0 ) ^{2}}{( 3 ) ^{2}} - \frac{( y - 0 ) ^{2}}{( \frac{3}{2} ) ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (0, 0), a = 3, b = \frac{3}{2}

\frac{( x - 3 ) ^{2}}{25} - \frac{( y + 2 ) ^{2}}{9} = 1

x^{2} - y^{2} - 8 x - 8 y - 15 = 0

\frac{a + 10}{b} = \frac{b}{a}

(x - 2)^{2} - (y + 5)^{2} = 18

\frac{4 y ^{2}}{9} - \frac{4 x ^{2}}{27} = 1

What is 4x^2-8y^2<= 12 ?
The solution to 4x^2-8y^2<= 12 is Hyperbola with (h,k)=(0,0),a=sqrt(3),b=sqrt(3/2)