What is the center ((x-3)^2}{16}+\frac{(y-4)^2)/9 =1 ?

The center ((x-3)^2}{16}+\frac{(y-4)^2)/9 =1 is (3,4)

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center ((x-3)^2}{16}+\frac{(y-4)^2)/9 =1

center

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 4 ) ^{2}}{9} = 1

(3, 4)

Solution steps

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 4 ) ^{2}}{9} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 4 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

And the center is:

(3, 4)

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

\frac{( x - 8 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{9} = 1

2 x^{2} + 2 y^{2} + 16 y = 0

x^{2} + 9 y^{2} + 6 x - 90 y + 225 = 0

13 x^{2} + 49 y^{2} = 637

What is the center ((x-3)^2}{16}+\frac{(y-4)^2)/9 =1 ?
The center ((x-3)^2}{16}+\frac{(y-4)^2)/9 =1 is (3,4)