Find the equation of the ellipse in standard form if: the latus rectum

1.	Find the equation of the ellipse in standard form if: the latus rectum has length 6 and foci are (±2, 0).
Answer» Given, the length of the latus rectum is 6, and co-ordinates of foci are (±2, 0). The foci of the ellipse are on the X-axis. Let the required equation of ellipse be \(\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1\) where a > b. Length of latus rectum = 2b²/a 2b²/a =6 b² = 3a …..(i) Co-ordinates of foci are (±ae, 0) ae = 2 a² e² = 4 …..(ii) Now, b² = a² (1 – e²) b² = a² – a² e² 3a = a² – 4 …..[From (i) and (ii)] a² – 3a – 4 = 0 a² – 4a + a – 4 = 0 a(a – 4) + 1(a – 4) = 0 (a – 4) (a + 1) = 0 a – 4 = 0 or a + 1 = 0 a = 4 or a = -1 Since a = -1 is not possible, a = 4 a² = 4 Substituting a = 4 in (i), we get b = 3(4) = 12 The required equation of ellipse is \(\frac {x^2}{16}+\frac{y^2}{12}=1\)

Find the equation of the ellipse in standard form if: the latus rectum has length 6 and foci are (±2, 0).

Answer»

Given, the length of the latus rectum is 6, and co-ordinates of foci are (±2, 0). The foci of the ellipse are on the X-axis.

Let the required equation of ellipse be \(\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1\) where a > b.

Length of latus rectum = 2b²/a

2b²/a =6

b² = 3a …..(i)

Co-ordinates of foci are (±ae, 0)

ae = 2

a² e² = 4 …..(ii)

Now, b² = a² (1 – e²)

b² = a² – a² e²

3a = a² – 4 …..[From (i) and (ii)]

a² – 3a – 4 = 0

a² – 4a + a – 4 = 0

a(a – 4) + 1(a – 4) = 0

(a – 4) (a + 1) = 0

a – 4 = 0 or a + 1 = 0

a = 4 or a = -1

Since a = -1 is not possible,

a = 4

a² = 4

Substituting a = 4 in (i), we get b = 3(4) = 12

The required equation of ellipse is \(\frac {x^2}{16}+\frac{y^2}{12}=1\)