What is the axis ((x-1)^2)/(25)+((y+3)^2)/(16)=1 ?

The axis ((x-1)^2)/(25)+((y+3)^2)/(16)=1 is semi-major axis a=5, semi-minor axis b=4

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axis ((x-1)^2)/(25)+((y+3)^2)/(16)=1

axis

\frac{( x - 1 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{16} = 1

semi - major axis a = 5, semi - minor axis b = 4

Solution steps

\frac{( x - 1 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 1 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (1, - 3), a = 5, b = 4

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (1, - 3), a = 5, b = 4

3 x^{2} + y^{2} - 24 x + 39 = 0

\frac{x ^{2}}{5} + \frac{y ^{2}}{9} = 1

36 x^{2} + 5 y^{2} - 90 y - 495 = 0

x^{2} + 4 y^{2} = 46

9 x^{2} + y^{2} = 9

What is the axis ((x-1)^2)/(25)+((y+3)^2)/(16)=1 ?
The axis ((x-1)^2)/(25)+((y+3)^2)/(16)=1 is semi-major axis a=5, semi-minor axis b=4