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

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

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4x^2+y^2<= 4

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

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

Solution steps

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

Rewrite

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

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^{2} + (y - 1)^{2} + (4 + y)^{2} = 64

\frac{( y + 1 ) ^{2}}{16} + \frac{x ^{2}}{4} - 1 = 0

\frac{( y - 4 ) ^{2}}{25} + \frac{( x + 2 ) ^{2}}{4} = 1

\frac{x ^{2}}{8 ^{2}} + \frac{y ^{2}}{5 ^{2}} = 1

x^{2} - 4 x + 4 y^{2} + 4 y + 1 = 0

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