What is the axis ((x+1)^2)/(64)+((y-6)^2)/(16)=1 ?

The axis ((x+1)^2)/(64)+((y-6)^2)/(16)=1 is semi-major axis a=8, semi-minor axis b=4

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axis ((x+1)^2)/(64)+((y-6)^2)/(16)=1

axis

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 6 ) ^{2}}{16} = 1

semi - major axis a = 8, semi - minor axis b = 4

Solution steps

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 6 ) ^{2}}{16} = 1

Rewrite

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 6 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 1 ) ) ^{2}}{8 ^{2}} + \frac{( y - 6 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 1, 6), a = 8, b = 4

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 1, 6), a = 8, b = 4

\frac{x ^{2}}{144} + \frac{y ^{2}}{81} = 1

16 x^{2} - 64 x + 9 y^{2} - 54 y = - 1

\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1

4 x^{2} + 9 y^{2} - 64 x + 54 y + 301 = 0

x^{2} + 4 y^{2} = 10

What is the axis ((x+1)^2)/(64)+((y-6)^2)/(16)=1 ?
The axis ((x+1)^2)/(64)+((y-6)^2)/(16)=1 is semi-major axis a=8, semi-minor axis b=4