# SOLUTION: Find the eccentricity of the ellipse given by {{{16x^2+25y^2=100}}} I have tried: {{{(16x^2)/100 + (25y^2)/100 = 1}}} that reduces to: {{{4x^2/25 + 1y^2/4 =1}}} W

Algebra ->  Algebra  -> Quadratic-relations-and-conic-sections -> SOLUTION: Find the eccentricity of the ellipse given by {{{16x^2+25y^2=100}}} I have tried: {{{(16x^2)/100 + (25y^2)/100 = 1}}} that reduces to: {{{4x^2/25 + 1y^2/4 =1}}} W      Log On

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 Algebra: Conic sections - ellipse, parabola, hyperbola Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Quadratic-relations-and-conic-sections Question 145168: Find the eccentricity of the ellipse given by I have tried: that reduces to: What do I do next? Please explain if you can! Thanks so very much!Answer by Edwin McCravy(8880)   (Show Source): You can put this solution on YOUR website!Find the eccentricity of the ellipse given by I have tried: that reduces to: What do I do next? Please explain if you can! Thanks so very much! ``` Your error is in thinking that you are necessarily reducing the fractions to lowest terms. Instead think of it as making the numerators 1. Sometimes it amounts to reducing the fraction but not always. Do it this way instead. Go back to: Get a coefficient where the is by multiplying top and bottom of the first fraction by , and get a coefficient where the is on the second fraction by multiplying top and bottom by , and we have this: Now cancel and that leaves just 1 understood on top: So we have And we only need to reduce the fractions on the bottom The larger denominator is , and since , we compare the above to: So Eccentricity of an ellipse = where Eccentricity of this elipse = Edwin```