What is 2x^2+y^2<= 8 ?

The solution to 2x^2+y^2<= 8 is Ellipse with (h,k)=(0,0),a=2,b=2sqrt(2)

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2x^2+y^2<= 8

2 x^{2} + y^{2} \leq 8

(h, k) = (0, 0), a = 2, b = 22

Solution steps

2 x^{2} + y^{2} \leq 8

Rewrite

2 x^{2} + y^{2} \leq 8

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 0), a = 2, b = 22

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 0), b = 22, a = 2

x^{2} + 4 y^{2} - 12 x - 16 y = 48

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

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

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

100 x^{2} + 4 y^{2} + 1200 x - 32 y + 3264 = 0

What is 2x^2+y^2<= 8 ?
The solution to 2x^2+y^2<= 8 is Ellipse with (h,k)=(0,0),a=2,b=2sqrt(2)