What is x^2+4y^2<= 4 ?

The solution to x^2+4y^2<= 4 is Ellipse with (h,k)=(0,0),a=2,b=1

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x^2+4y^2<= 4

x^{2} + 4 y^{2} \leq 4

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

Solution steps

x^{2} + 4 y^{2} \leq 4

Rewrite

x^{2} + 4 y^{2} \leq 4

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

2 x^{2} + 4 y^{2} + 10 x + 9 y - 5 = 0

\frac{( y - 6 ) ^{2}}{36} + \frac{( x + 3 ) ^{2}}{16} = 1

\frac{x ^{2}}{9} + \frac{( y - 4 ) ^{2}}{7} = 1

(x + 4)^{2} + (\frac{y - 3}{4})^{2} = 1

\frac{( x - 1 ) ^{2}}{5} + \frac{( y - 3 ) ^{2}}{4} = 1

What is x^2+4y^2<= 4 ?
The solution to x^2+4y^2<= 4 is Ellipse with (h,k)=(0,0),a=2,b=1