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

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

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2x^2+y^2<= 2

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

(h, k) = (0, 0), a = 1, b = 2

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 0), a = 1, b = 2

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 0), b = 2, a = 1

(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

16 x^{2} + 15 y^{2} - 64 x - 90 y - 41 = 0

25 y^{2} + 9 x^{2} - 108 x - 200 y + 499 = 0

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