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

The eccentricity 41x^2+16y^2+246x-192y+289=0 is 5/(sqrt(41))

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

eccentricity

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

\frac{5}{41}

Solution steps

= \frac{b ^{2} - a ^{2}}{b}

Calculate ellipse properties

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

Ellipse with center

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

= \frac{( 41 ) ^{2} - 4 ^{2}}{41}

\frac{( 41 ) ^{2} - 4 ^{2}}{41} = \frac{5}{41}

= \frac{5}{41}

\frac{x ^{2}}{9} + \frac{y ^{2}}{25} = 1

25 x^{2} + 16 y^{2} = 400

x^{2} + 4 y^{2} - 6 x - 16 y + 21 = 0

\frac{x ^{2}}{16} + \frac{y ^{2}}{12} = 1

x^{2} + 3 y^{2} = 1

What is the eccentricity 41x^2+16y^2+246x-192y+289=0 ?
The eccentricity 41x^2+16y^2+246x-192y+289=0 is 5/(sqrt(41))