What is x^2+5y^2=2021y ?

The solution to x^2+5y^2=2021y is Ellipse with (h,k)=(0, 2021/10),a=(2021)/(2sqrt(5)),b= 2021/10

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x^2+5y^2=2021y

x^{2} + 5 y^{2} = 2021 y

(h, k) = (0, \frac{2021}{10}), a = \frac{2021}{2 5}, b = \frac{2021}{10}

Solution steps

x^{2} + 5 y^{2} = 2021 y

Rewrite

x^{2} + 5 y^{2} = 2021 y

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( \frac{2021}{2 5} ) ^{2}} + \frac{( y - \frac{2021}{10} ) ^{2}}{( \frac{2021}{10} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, \frac{2021}{10}), a = \frac{2021}{2 5}, b = \frac{2021}{10}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, \frac{2021}{10}), a = \frac{2021}{2 5}, b = \frac{2021}{10}

(x + y) - x^{2} = x^{2} + y^{2}

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

9 x^{2} + 16 y^{2} = 4

(x)^{2} + \frac{( y - 6 ) ^{2}}{12} = 1

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

What is x^2+5y^2=2021y ?
The solution to x^2+5y^2=2021y is Ellipse with (h,k)=(0, 2021/10),a=(2021)/(2sqrt(5)),b= 2021/10