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प्रश्न
Find the
- lengths of the principal axes.
- co-ordinates of the focii
- equations of directrices
- length of the latus rectum
- distance between focii
- distance between directrices of the ellipse:
3x2 + 4y2 = 1
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उत्तर
The equation of the ellipse is 3x2 + 4y2 = 1
i.e., `x^2/((1/3)) + y^2/((1/4))` = 1
Comparing with `x^2/"a"^2 + y^2/"b"^2` = 1, we get
a2 = `1/3`, b2 = `1/4`
∴ a = `1/sqrt3`, b = `1/2`
∴ a > b
i. Length of major axis = 2a = `2/sqrt(3)`
Length of minor axis = 2b = `2(1/2)` = 1
ii. Eccentricity = e = `sqrt("a"^2 - "b"^2)/"a"`
= `sqrt(1/3 - 1/4)/((1/sqrt(3))`
= `((1/(2sqrt(3))))/((1/sqrt(3))) = 1/2`
∴ ae = `1/sqrt(3) xx 1/2 = 1/(2sqrt(3))`
∴ coordinates of foci = (± ae, 0) = `(± 1/(2sqrt(3)), 0)`.
iii. `"a"/"e" = ((1/sqrt(3)))/((1/2)) = 2/sqrt(3)`
The equations of directrices are
x = `± "a"/"e"`
∴ x = `± 2/sqrt(3)`
iv. Length of latus rectum = `(2"b"^2)/"a"`
= `(2 xx 1/4)/((1/sqrt(3))`
= `sqrt(3)/2`
v. Distance between foci = 2ae
= `2 xx 1/(2sqrt(3)`
= `1/sqrt(3)`
vi. Distance between directrices = `(2"a")/"e"`
= `2 xx 2/sqrt(3)`
= `4/sqrt(3)`
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