What is the axis 81x^2-1296x+16y^2+64y=-3952 ?

The axis 81x^2-1296x+16y^2+64y=-3952 is semi-major axis b=9, semi-minor axis a=4

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axis 81x^2-1296x+16y^2+64y=-3952

axis

semi - major axis b = 9, semi - minor axis a = 4

Solution steps

81 x^{2} - 1296 x + 16 y^{2} + 64 y = - 3952

Rewrite

81 x^{2} - 1296 x + 16 y^{2} + 64 y = - 3952

in the form of the standard ellipse equation

\frac{( x - 8 ) ^{2}}{4 ^{2}} + \frac{( y - ( - 2 ) ) ^{2}}{9 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (8, - 2), a = 4, b = 9

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (8, - 2), b = 9, a = 4

What is the axis 81x^2-1296x+16y^2+64y=-3952 ?
The axis 81x^2-1296x+16y^2+64y=-3952 is semi-major axis b=9, semi-minor axis a=4