What is the center ((x-5)^2)/(49)+((y-9)^2)/(25)=1 ?

The center ((x-5)^2)/(49)+((y-9)^2)/(25)=1 is (5,9)

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center ((x-5)^2)/(49)+((y-9)^2)/(25)=1

center

\frac{( x - 5 ) ^{2}}{49} + \frac{( y - 9 ) ^{2}}{25} = 1

(5, 9)

Solution steps

\frac{( x - 5 ) ^{2}}{49} + \frac{( y - 9 ) ^{2}}{25} = 1

Rewrite

\frac{( x - 5 ) ^{2}}{49} + \frac{( y - 9 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

\frac{( x - 5 ) ^{2}}{7 ^{2}} + \frac{( y - 9 ) ^{2}}{5 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (5, 9), a = 7, b = 5

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (5, 9), a = 7, b = 5

And the center is:

(5, 9)

4 x^{2} + 9 y^{2} = 25

3 x^{(2)} + 2 y^{(2)} - 12 x + 8 y + 19 = 0

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

3 x^{2} + y^{2} + 2 x - 6 y = 3

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

What is the center ((x-5)^2)/(49)+((y-9)^2)/(25)=1 ?
The center ((x-5)^2)/(49)+((y-9)^2)/(25)=1 is (5,9)