What is the center 41x^2+16y^2+246x-192y+289=0 ?

The center 41x^2+16y^2+246x-192y+289=0 is (-3,6)

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center 41x^2+16y^2+246x-192y+289=0

center

41 x^{2} + 16 y^{2} + 246 x - 192 y + 289 = 0

(- 3, 6)

Solution steps

41 x^{2} + 16 y^{2} + 246 x - 192 y + 289 = 0

Rewrite

41 x^{2} + 16 y^{2} + 246 x - 192 y + 289 = 0

in the form of the standard ellipse equation

\frac{( x - ( - 3 ) ) ^{2}}{4 ^{2}} + \frac{( y - 6 ) ^{2}}{( 41 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 3, 6), a = 4, b = 41

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 3, 6), b = 41, a = 4

And the center is:

(- 3, 6)

x^{2} + 14 x + y^{2} - 6 y - 23 = 0

x = 2 y - y^{2}

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

x = 2 y^{2}

f (x) = - x^{2}

What is the center 41x^2+16y^2+246x-192y+289=0 ?
The center 41x^2+16y^2+246x-192y+289=0 is (-3,6)