What is the center ((x-3)^2)/(16)+((y+1)^2)/(36)=1 ?

The center ((x-3)^2)/(16)+((y+1)^2)/(36)=1 is (3,-1)

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center ((x-3)^2)/(16)+((y+1)^2)/(36)=1

center

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 1 ) ^{2}}{36} = 1

(3, - 1)

Solution steps

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 1 ) ^{2}}{36} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 1 ) ^{2}}{36} = 1

in the form of the standard ellipse equation

\frac{( x - 3 ) ^{2}}{4 ^{2}} + \frac{( y - ( - 1 ) ) ^{2}}{6 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, - 1), a = 4, b = 6

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, - 1), b = 6, a = 4

And the center is:

(3, - 1)

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

5 x^{2} + 9 y^{2} - 10 x + 4 = 0

3 x^{2} + 4 y^{2} = 12

\frac{x ^{2}}{7} + \frac{y ^{2}}{16} = 1

25 (x - 2)^{2} + 9 (y + 3)^{2} = 225

What is the center ((x-3)^2)/(16)+((y+1)^2)/(36)=1 ?
The center ((x-3)^2)/(16)+((y+1)^2)/(36)=1 is (3,-1)