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प्रश्न
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
उत्तर
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`
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