What is 2x^2+y^2<= 1 ?

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

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2x^2+y^2<= 1

2 x^{2} + y^{2} \leq 1

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

Solution steps

2 x^{2} + y^{2} \leq 1

Rewrite

2 x^{2} + y^{2} \leq 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

\frac{x ^{2}}{7 ^{2}} + \frac{y ^{2}}{3 ^{2}} = 1

16 x^{2} + 25 y^{2} - 64 x + 150 y = - 279

4 x^{2} + y^{2} - 40 x + 4 y + 68 = 0

x^{2} + 2 x = - 4 y^{2} + 12 y + 7

x^{2} + 3 y^{2} + 10 x + 9 y - 20 = 0

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