x^2+2y^2<1

Solution

x^{2} + 2 y^{2} < 1

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

Solution steps

x^{2} + 2 y^{2} < 1

Rewrite

x^{2} + 2 y^{2} < 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

(3 x - 3)^{2} + (2 y - 4)^{2} = 43 + 9 + 16

x^{2} + y^{2} + 12 y = 0

9 x^{2} + 25 y^{2} + 108 x - 200 y + 4 = 0

9 x^{2} + 16 y^{2} + 18 x - 32 y - 119 = 0

\frac{x ^{2}}{16} + \frac{( y - 1 ) ^{2}}{36} = 1

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