What is 20x^2+4y^2=5 ?

The solution to 20x^2+4y^2=5 is Ellipse with (h,k)=(0,0),a= 1/2 ,b=(sqrt(5))/2

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20x^2+4y^2=5

20 x^{2} + 4 y^{2} = 5

(h, k) = (0, 0), a = \frac{1}{2}, b = \frac{5}{2}

Solution steps

20 x^{2} + 4 y^{2} = 5

Rewrite

20 x^{2} + 4 y^{2} = 5

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( \frac{1}{2} ) ^{2}} + \frac{( y - 0 ) ^{2}}{( \frac{5}{2} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = \frac{1}{2}, b = \frac{5}{2}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 0), b = \frac{5}{2}, a = \frac{1}{2}

\frac{( x - 5 ) ^{2}}{12} + \frac{( y + 2 ) ^{2}}{16} = 1

\frac{( x - 2 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{9} = 1

x^{2} + 2 y^{2} - 10 x + 8 y + 29 = 0

16 x^{2} + 4 y^{2} - 32 x + 16 y - 32 = 0

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

What is 20x^2+4y^2=5 ?
The solution to 20x^2+4y^2=5 is Ellipse with (h,k)=(0,0),a= 1/2 ,b=(sqrt(5))/2