What is the center ((x-6)^2)/5+((y+1)^2)/6 =1 ?

The center ((x-6)^2)/5+((y+1)^2)/6 =1 is (6,-1)

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center ((x-6)^2)/5+((y+1)^2)/6 =1

center

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

(6, - 1)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (6, - 1), a = 5, b = 6

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (6, - 1), b = 6, a = 5

And the center is:

(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

x^{2} + 5 y^{2} - 4 x + 20 y = - 19

What is the center ((x-6)^2)/5+((y+1)^2)/6 =1 ?
The center ((x-6)^2)/5+((y+1)^2)/6 =1 is (6,-1)