What is the axis ((x-5)^2)/4+((y+3)^2)/(16)=1 ?

The axis ((x-5)^2)/4+((y+3)^2)/(16)=1 is semi-major axis b=4, semi-minor axis a=2

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axis ((x-5)^2)/4+((y+3)^2)/(16)=1

axis

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{16} = 1

semi - major axis b = 4, semi - minor axis a = 2

Solution steps

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - 5 ) ^{2}}{2 ^{2}} + \frac{( y - ( - 3 ) ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (5, - 3), a = 2, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (5, - 3), b = 4, a = 2

x^{2} + 2 y^{2} + 2 x - 8 y + 7 = 0

\frac{x ^{2}}{36} + \frac{y ^{2}}{11} = 1

36 x^{2} + 11 y^{2} + 144 x + 44 y - 208 = 0

9 x^{2} + 16 y^{2} + 90 x + 32 y + 97 = 0

\frac{( x - 4 ) ^{2}}{4} + \frac{y ^{2}}{16} = 1

What is the axis ((x-5)^2)/4+((y+3)^2)/(16)=1 ?
The axis ((x-5)^2)/4+((y+3)^2)/(16)=1 is semi-major axis b=4, semi-minor axis a=2