Find the standard form of the equation of the ellipse with the given characteristics?

Foci: (0, 0) and (4, 0)
Major Axis of length 8

2 Answers
Shwetank Mauria
Jul 4, 2018

Equation is (x-2)^2/16+y^2/12=1(x−2)216+y212=1

Explanation:

As focii are (0,0)(0,0) and (4,0)(4,0), center of ellipse is midpoint i.e. (2,0)(2,0) and major axis is 88, equation is of the form

(x-2)^2/4^2+(y-0)^2/b^2=1(x−2)242+(y−0)2b2=1

where bb is half minor axis.

As distance between focii is 44 and major axis is 88, eccentricity is 4/8=1/248=12 and

(1/2)^2=1-b^2/4^2(12)2=1−b242

or b^2/16=1-1/4=3/4b216=1−14=34

and b^2=12b2=12

Hence equation of ellipse is
(x-2)^2/16+(y-0)^2/12=1(x−2)216+(y−0)212=1

or (x-2)^2/16+y^2/12=1(x−2)216+y212=1

Shwetank Mauria
Jul 4, 2018

Equation is (x-2)^2/16+y^2/12=1(x−2)216+y212=1

Explanation:

As focii are (0,0)(0,0) and (4,0)(4,0), center of ellipse is midpoint i.e. (2,0)(2,0) and major axis is 88, equation is of the form

(x-2)^2/4^2+(y-0)^2/b^2=1(x−2)242+(y−0)2b2=1

where bb is half minor axis.

As distance between focii is 44 and major axis is 88, eccentricity is 4/8=1/248=12 and

(1/2)^2=1-b^2/4^2(12)2=1−b242

or b^2/16=1-1/4=3/4b216=1−14=34

and b^2=12b2=12

Hence equation of ellipse is
(x-2)^2/16+(y-0)^2/12=1(x−2)216+(y−0)212=1

or (x-2)^2/16+y^2/12=1(x−2)216+y212=1

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