What is the axis ((x-4)^2)/(64)+((y-2)^2)/(39)=1 ?

The axis ((x-4)^2)/(64)+((y-2)^2)/(39)=1 is semi-major axis a=8, semi-minor axis b=sqrt(39)

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axis ((x-4)^2)/(64)+((y-2)^2)/(39)=1

axis

\frac{( x - 4 ) ^{2}}{64} + \frac{( y - 2 ) ^{2}}{39} = 1

semi - major axis a = 8, semi - minor axis b = 39

Solution steps

\frac{( x - 4 ) ^{2}}{64} + \frac{( y - 2 ) ^{2}}{39} = 1

Rewrite

\frac{( x - 4 ) ^{2}}{64} + \frac{( y - 2 ) ^{2}}{39} = 1

in the form of the standard ellipse equation

\frac{( x - 4 ) ^{2}}{8 ^{2}} + \frac{( y - 2 ) ^{2}}{( 39 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (4, 2), a = 8, b = 39

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (4, 2), a = 8, b = 39

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What is the axis ((x-4)^2)/(64)+((y-2)^2)/(39)=1 ?
The axis ((x-4)^2)/(64)+((y-2)^2)/(39)=1 is semi-major axis a=8, semi-minor axis b=sqrt(39)