Question

Find the

1. lengths of the principal axes.
2. co-ordinates of the focii 
3. equations of directrics 
4. length of the latus rectum
5. distance between focii 
6. distance between directrices of the ellipse:

2x² + 6y² = 6

Answer 1

Given equation of the ellipse is 2x² + 6y² = 6.

∴ `x^2/3 + y^2/1` = 1

Comparing this equation with `x^2/"a"^2 + y^2/"b"^2` = 1, we get

a² = 3 and b² = 1

∴ a = `sqrt(3) and "b"` = 1

Since a > b,

X-axis is the major axis and Y-axis is the minor axis.

i. Length of major axis = 2a = `2sqrt(3)`

Length of minor axis = 2b = 2(1) = 2

∴ Lengths of the principal axes are `2sqrt(3)` and 2.

ii. We know that e = `sqrt("a"^2 - "b"^2)/"a"`

∴ e = `sqrt(3 - 1)/sqrt(3) = sqrt(2)/sqrt(3)`

Co-ordinates of the foci are S(ae, 0) and S'(–ae, 0),

i.e., `"S"(sqrt(3)(sqrt(2)/sqrt(3)),0)` and `"S""'"(-sqrt(3)(sqrt(2)/sqrt(3)),0)`,

i.e., `"S"(sqrt(2), 0)` and `"S""'"(-sqrt(2), 0)`

iii. Equations of the directrices are x = `±"a"/"e"`,

i.e., x = `±sqrt3/(sqrt2/sqrt3)`, i.e., x = `±3/sqrt(2)`

iv. Length of latus rectum = `(2"b"^2)/"a" = (2(1)^2)/sqrt(3) = 2/sqrt(3)`

v. Distance between foci = 2ae

= `2(sqrt(3)) (sqrt(2)/sqrt(3))`

= `2sqrt(2)`

vi. Distance between directrices = `(2"a")/"e"`

= `(2sqrt(3))/(sqrt(2)/sqrt(3))`

= `(2 xx 3)/sqrt(2)`

= `3sqrt(2)`