What is the center ((x-1)^2)/6+((y-3)^2)/2 =1 ?

The center ((x-1)^2)/6+((y-3)^2)/2 =1 is (1,3)

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center ((x-1)^2)/6+((y-3)^2)/2 =1

center

\frac{( x - 1 ) ^{2}}{6} + \frac{( y - 3 ) ^{2}}{2} = 1

(1, 3)

Solution steps

\frac{( x - 1 ) ^{2}}{6} + \frac{( y - 3 ) ^{2}}{2} = 1

Rewrite

\frac{( x - 1 ) ^{2}}{6} + \frac{( y - 3 ) ^{2}}{2} = 1

in the form of the standard ellipse equation

\frac{( x - 1 ) ^{2}}{( 6 ) ^{2}} + \frac{( y - 3 ) ^{2}}{( 2 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (1, 3), a = 6, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (1, 3), a = 6, b = 2

And the center is:

(1, 3)

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

25 x^{2} + 16 y^{2} = 400

x^{2} + y^{2} - 10 y = 0

\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1

\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1

What is the center ((x-1)^2)/6+((y-3)^2)/2 =1 ?
The center ((x-1)^2)/6+((y-3)^2)/2 =1 is (1,3)