What is the center (((x-2)^2}{16})+(\frac{y^2)/9)=1 ?

The center (((x-2)^2}{16})+(\frac{y^2)/9)=1 is (2,0)

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

center

(2, 0)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - 2 ) ^{2}}{4 ^{2}} + \frac{( y - 0 ) ^{2}}{3 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (2, 0), a = 4, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (2, 0), a = 4, b = 3

And the center is:

(2, 0)

What is the center (((x-2)^2}{16})+(\frac{y^2)/9)=1 ?
The center (((x-2)^2}{16})+(\frac{y^2)/9)=1 is (2,0)