Question

Arrange the equations of ellipses in increasing order of their eccentricities

Answer 1

we know that

The equation for the eccentricity of an ellipse is

e=c/a

where

e is eccentricity, c is the distance from the foci to the center, and a is the square root of the larger of our two denominators.

To find c, we must use the equation a²−b²=c², where b is the square root of the smaller of our two denominators

case 1)
(x²/2²)+(y-2)²/4²=1
a=4
b=2
c²=16-4----> c=√12
e=c/a----> √12/4----> e=0.8660

case 2)
(x+3)²/5²+(y-1)²/3²=1
a=5
b=3
c²=25-9----> c=√16-----> c=4
e=c/a----> 4/5-----> e=0.80

case 3)
(x-5)²/3²+(y²/7²)=1
a=7
b=3
c²=49-9-----> c=√40
e=c/a-----> √40/7-----> e=0.9035

case 4)
(x-2)²/4²+(y+4)²7²=1
a=7
b=4
c²=49-16-----> c=√33
e=c/a-----> √33/7------> e=0.8207

case 5)
x²/7²+y²/6²=1
a=7
b=6
c²=49-36-----> c=√13
e=c/a-----> √13/7-----> e=0.5151

case 6)
(x-3)²/6²+y²/4²=1
a=6
b=4
c²=36-16-----> c=√20
e=c/a-----> √20/6-----> e=0.7454

case 7)
(x+4)²/5²+(y-5)²/6²=1
a=6
b=5
c²=36-25-----> c=√11
e=c/a-----> √11/6----> e=0.5528

case 8)
x²/7²+(y+7)²/2²=1
a=7
b=2
c²=49-4----> c=√45
e=c/a-----> √45/7----> e=0.9583

the answer is
1) x²/7²+y²/6²=1 e=0.5151
2) (x+4)²/5²+(y-5)²/6²=1 e=0.5528
3) (x-3)²/6²+y²/4²=1 e=0.7454
4) (x+3)²/5²+(y-1)²/3²=1 e=0.8000
5) (x-2)²/4²+(y+4)²7²=1 e=0.8207
6) (x²/2²)+(y-2)²/4²=1 e=0.8660
7) (x+4)²/5²+(y-5)²/6²=1 e=0.9035
8) x²/7²+(y+7)²/2²=1 e=0.9583