What is 25x^2+(y^2-4y)=21 ?

The solution to 25x^2+(y^2-4y)=21 is Ellipse with (h,k)=(0,2),a=1,b=5

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25x^2+(y^2-4y)=21

25 x^{2} + (y^{2} - 4 y) = 21

(h, k) = (0, 2), a = 1, b = 5

Solution steps

25 x^{2} + (y^{2} - 4 y) = 21

Rewrite

25 x^{2} + (y^{2} - 4 y) = 21

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 2), a = 1, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 2), b = 5, a = 1

\frac{x ^{2}}{25} - 13 y + 17 x = - \frac{y ^{2}}{49} + 1

x, \frac{Y ( x ) ^{2}}{4 - x ^{2}} = 1

2 x^{2} + 10 y^{2} - 10 x - 6 y - 10 = 0

\frac{y ^{2}}{16} = 1 - \frac{x ^{2}}{9}

4 x^{2} + 9 y^{2} + 16 x - 36 y + 16 = 0

What is 25x^2+(y^2-4y)=21 ?
The solution to 25x^2+(y^2-4y)=21 is Ellipse with (h,k)=(0,2),a=1,b=5