What is the center ((x-4)^2)/(3^2)+((y-2)^2)/(4^2)=1 ?

The center ((x-4)^2)/(3^2)+((y-2)^2)/(4^2)=1 is (4,2)

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center ((x-4)^2)/(3^2)+((y-2)^2)/(4^2)=1

center

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

(4, 2)

Solution steps

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

And the center is:

(4, 2)

4 y^{2} + 2 x^{2} - 4 y - 8 x - 15 = 0

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

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

36 x^{2} + 9 y^{2} + 48 x - 36 y + 43 = 0

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

What is the center ((x-4)^2)/(3^2)+((y-2)^2)/(4^2)=1 ?
The center ((x-4)^2)/(3^2)+((y-2)^2)/(4^2)=1 is (4,2)