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Question
Answer the following:
Find the
- lengths of the principal axes
- co-ordinates of the foci
- equations of directrices
- length of the latus rectum
- distance between foci
- distance between directrices of the ellipse:
`x^2/25 + y^2/9` = 1
Solution
Given equation of the ellipse is `x^2/25 + y^2/9` = 1.
Comparing this equation with `x^2/"a"^2 + y^2/"b"^2` = 1, we get
a2 = 25 and b2 = 9
∴ a = 5 and b = 3
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(3) = 6
∴ Lengths of the principal axes are 10 and 6.
ii. We know that e = `sqrt("a"^2 - "b"^2)/"a"`
∴ e = `sqrt(25- 9)/5`
= `sqrt(16)/5`
= `4/5`
Co-ordinates of the foci are S(ae, 0) and S'(–ae, 0),
i.e., `"S"(5(4/5),0)` and `"S'"(-5(4/5),0)`,
i.e., S(4, 0) and S'(–4, 0)
iii. Equations of the directrices are x = `± "a"/"e"`,
i.e., x = `± 5/(4/5)`, i.e., x = `± 25/4`
iv. Length of latus rectum = `(2"b"^2)/"a" = (2(3)^2)/5 = 18/5`
v. Distance between foci = 2ae = `2(5)(4/5)` = 8
vi. Distance between directrices = `(2"a")/"e"`
= `(2(5))/(4/5)`
= `25/2`
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Find the
- lengths of the principal axes.
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- equations of directrics
- length of the latus rectum
- distance between focii
- distance between directrices of the ellipse:
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