What is the axis ((x+1)^2)/9+((y-3)^2)/(16)=1 ?

The axis ((x+1)^2)/9+((y-3)^2)/(16)=1 is semi-major axis b=4, semi-minor axis a=3

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axis ((x+1)^2)/9+((y-3)^2)/(16)=1

axis

\frac{( x + 1 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

semi - major axis b = 4, semi - minor axis a = 3

Solution steps

\frac{( x + 1 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

Rewrite

\frac{( x + 1 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 1, 3), a = 3, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 1, 3), b = 4, a = 3

\frac{x ^{2}}{64} + \frac{y ^{2}}{16} = 1

\frac{x ^{2}}{9} + \frac{y ^{2}}{64} = 1

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

\frac{x ^{2}}{3} + \frac{y ^{2}}{2} = 1

36 x^{2} + 9 y^{2} - 324 = 0

What is the axis ((x+1)^2)/9+((y-3)^2)/(16)=1 ?
The axis ((x+1)^2)/9+((y-3)^2)/(16)=1 is semi-major axis b=4, semi-minor axis a=3