What is x^2+16y^2-128y+240=0 ?

The solution to x^2+16y^2-128y+240=0 is Ellipse with (h,k)=(0,4),a=4,b=1

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x^2+16y^2-128y+240=0

x^{2} + 16 y^{2} - 128 y + 240 = 0

(h, k) = (0, 4), a = 4, b = 1

Solution steps

x^{2} + 16 y^{2} - 128 y + 240 = 0

Rewrite

x^{2} + 16 y^{2} - 128 y + 240 = 0

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{4 ^{2}} + \frac{( y - 4 ) ^{2}}{1 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 4), a = 4, b = 1

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 4), a = 4, b = 1

4 x^{2} + y^{2} = 6 y - 5

18 x^{2} + 12 y^{2} + 24 x - 12 y = 1

16 x^{2} + 25 y^{2} + 32 x - 150 y - 159 = 0

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

4 x^{2} - 8 x + 9 y^{2} - 72 y + 112 = 0

What is x^2+16y^2-128y+240=0 ?
The solution to x^2+16y^2-128y+240=0 is Ellipse with (h,k)=(0,4),a=4,b=1