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 ->  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


   



Question 145168: Find the eccentricity of the ellipse given by
16x%5E2%2B25y%5E2=100

I have tried:
%2816x%5E2%29%2F100+%2B+%2825y%5E2%29%2F100+=+1

that reduces to:
4x%5E2%2F25+%2B+1y%5E2%2F4+=1
What do I do next? Please explain if you can! Thanks so very much!

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Find the eccentricity of the ellipse given by
16x%5E2%2B25y%5E2=100

I have tried:
%2816x%5E2%29%2F100+%2B+%2825y%5E2%29%2F100+=+1

that reduces to:
4x%5E2%2F25+%2B+1y%5E2%2F4+=1
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:

16x%5E2%2F100+%2B+25y%5E2%2F100+=+1

Get a 1 coefficient where the 16 is by 
multiplying top and bottom of the first fraction by
%281%2F16%29, and get a 1 coefficient where 
the 25 is on the second fraction by multiplying 
top and bottom by %281%2F25%29, and we have this:



Now cancel and that leaves just 1 understood on top:



So we have

x%5E2+%2F%28100%2F16%29+%2B+y%5E2%2F%28100%2F25%29+=+1

And we only need to reduce the fractions on the bottom

x%5E2+%2F%2825%2F4%29+%2B+y%5E2%2F4+=+1

The larger denominator is a%5E2, and since 25%2F4%3E4,

we compare the above to:

x%5E2%2Fa%5E2+%2B+y%5E2%2Fb%5E2=1

So 

a%5E2=25%2F4            b%5E2=4
a=sqrt%2825%2F4%29        b=2 
a=5%2F2

Eccentricity of an ellipse = c%2Fa where c%5E2=a%5E2-b%5E2

c%5E2=a%5E2-b%5E2
c%5E2=%285%2F2%29%5E2-2%5E2
c%5E2=25%2F4-4
c%5E2=25%2F4-4%2F1
c%5E2=25%2F4-%284%2A4%29%2F%281%2A4%29
c%5E2=25%2F4-16%2F4
c%5E2=%2825-16%29%2F4
c%5E2=9%2F4
c=sqrt%289%2F4%29
c=3%2F2

Eccentricity of this elipse = 
  
Edwin