Find the eccentricity of the ellipse?

Question

the question is 16x^2 + 25y^2 = 100 and
my answer is e= .48 or 12/25
Can anyone verify this to be correct?

Answer 1

Urs is wrong.Here is the correct answer.------>
Given equation, 16 x^2 +25 y^2= 100

=> x^2/(100/16) +y^2/(100/25)=1
We know,
e= square root over[ 1-b^2/a^2]

so, here b=100/25 and a=100/16
Therefore, e=sq root over[ 1-(16/25)]
after simplification e=3/5 or 0.6.

Answer 2

e = sqrt[1 - b^2/a^2]

Where a is the semimajor axis length

and b is the semiminor axis length.

given the equation above, put it in the form


x^2/a^2 + y^2/b^2 = 1

then given the equation above you get:

16/100 x^2 + 25/100 y^2 = 1

1/a^2 = 16/100 so a^2 = 100/16 = 25/4

1/b^2 = 25/100 so b^2 = 100/25 = 4/1

e = SQRT[1 - (4/1)/(25/4)]

e = SQRT[1 - (4/1)(4/25)]

e = SQRT[1-16/25]

e = SQRT[9/25] = 3/5 = 0.60

QED

Answer 3

16x^2 + 25y^2 = 100
((x^2)/(25/4)) + ((y^2)/4) = 100

Eccentricity = (distance between the foci)/(length of major axis)

Foci = sqrt(a^2 - b^2)
Foci = sqrt((25/4) - 4)
Foci = sqrt((25 - 16)/4)
Foci = sqrt(9/4)
Foci = (3/2)

Distance between the Foci = 2(3/2) = 3

Major axis is sqrt(25/4) = (5/2)
Major Axis Length = 2(5/2) = 5

e = distance/length
e = (3/5)
e = .6