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

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

(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)

\frac{( x - 6 ) ^{2}}{5} + \frac{( y + 1 ) ^{2}}{6} = 1

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

x^{2} + 4 y^{2} = 36

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

4 x^{2} + y^{2} + 12 x + 6 y + 5 = 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)