Question

Answer the following:

Find the
(i) lengths of the principal axes
(ii) co-ordinates of the foci
(iii) equations of directrices
(iv) length of the latus rectum
(v) Distance between foci
(vi) distance between directrices of the curve

16x² + 25y² = 400

Answer 1

Given equation of the ellipse is 16x² + 25y² = 400

∴ `x^2/25 + y^2/16` = 1

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

a² = 25 and b² = 16

∴ a = 5 and b = 4

Since a > b,

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

i. Length of major axis = 2a = 2(5) = 10

Length of minor axis = 2b = 2(4) = 8

∴ Lengths of the principal axes are 10 and 8.

ii. b² = a²(1 – e²)

∴ 16 = 25(1 – e²)

∴ `16/25` = 1 – e² 

 ∴ e² = `1 - 16/25`

∴ e² = `9/25`

∴ e² = `3/5`  ...[∵ 0 < e < 1]

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

i.e., `"S"(5(3/5),0)` and `"S'"(-5(3/5),0)`,

i.e., S(3, 0) and S'(–3, 0)

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

i.e., x = `± 5/((3/5))`, i.e., x = `± 25/3`

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

= `(2(16))/5`

= `32/5`

v. Distance between foci = 2ae = `2(5)(3/5)` = 6

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

= `(2(5))/((3/5))`

= `50/3`