What is 5y(y+4x)+4x^2=-20(1-x)y+5 ?

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

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5y(y+4x)+4x^2=-20(1-x)y+5

5 y (y + 4 x) + 4 x^{2} = - 20 (1 - x) y + 5

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

Solution steps

5 y (y + 4 x) + 4 x^{2} = - 20 (1 - x) y + 5

Rewrite

5 y (y + 4 x) + 4 x^{2} + 20 (1 - x) y - 5 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

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

\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

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