What is 22x^2+4y^2=1 ?

The solution to 22x^2+4y^2=1 is Ellipse with (h,k)=(0,0),a=sqrt(1/22),b= 1/2

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22x^2+4y^2=1

22 x^{2} + 4 y^{2} = 1

(h, k) = (0, 0), a = \frac{1}{22}, b = \frac{1}{2}

Solution steps

22 x^{2} + 4 y^{2} = 1

Rewrite

22 x^{2} + 4 y^{2} = 1

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( \frac{1}{22} ) ^{2}} + \frac{( y - 0 ) ^{2}}{( \frac{1}{2} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = \frac{1}{22}, b = \frac{1}{2}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 0), b = \frac{1}{2}, a = \frac{1}{22}

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

16 x^{2} + 81 y^{2} = 81

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

\frac{( x + 1 ) ^{2}}{5} + \frac{( y - 3 ) ^{2}}{9} = 1

16 - 2 x^{2} - y^{2} > 0

What is 22x^2+4y^2=1 ?
The solution to 22x^2+4y^2=1 is Ellipse with (h,k)=(0,0),a=sqrt(1/22),b= 1/2