What is 4-12x^2-36y^2>= 0 ?

The solution to 4-12x^2-36y^2>= 0 is Ellipse with (h,k)=(0,0),a=sqrt(1/3),b= 1/3

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4-12x^2-36y^2>= 0

4 - 12 x^{2} - 36 y^{2} \geq 0

(h, k) = (0, 0), a = \frac{1}{3}, b = \frac{1}{3}

Solution steps

4 - 12 x^{2} - 36 y^{2} \geq 0

Rewrite

4 - 12 x^{2} - 36 y^{2} \geq 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 0), a = \frac{1}{3}, b = \frac{1}{3}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = \frac{1}{3}, b = \frac{1}{3}

x^{2} + 4 y^{2} + 4 x + 8 y - 8 = 0

x^{2} + 6 y^{2} = 15

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

(2 x + 4 y)^{2} = 16 x y + 16

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

What is 4-12x^2-36y^2>= 0 ?
The solution to 4-12x^2-36y^2>= 0 is Ellipse with (h,k)=(0,0),a=sqrt(1/3),b= 1/3