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प्रश्न
Find the
- lengths of the principal axes.
- co-ordinates of the focii
- equations of directrics
- length of the latus rectum
- distance between focii
- distance between directrices of the ellipse:
2x2 + 6y2 = 6
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उत्तर
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)`
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संबंधित प्रश्न
Find the
- lengths of the principal axes.
- co-ordinates of the focii
- equations of directrics
- length of the latus rectum
- distance between focii
- distance between directrices of the ellipse:
3x2 + 4y2 = 12
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- co-ordinates of the focii
- equations of directrices
- length of the latus rectum
- distance between focii
- distance between directrices of the ellipse:
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