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प्रश्न
Find the equation of the ellipse in standard form if the distance between foci is 6 and the distance between directrix is `50/3`.
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उत्तर
Let the required equation of ellipse be `x^2/"a"^2 + y^2/"b"^2` = 1, where a > b.
Distance between foci = 2ae
Given, distance between foci = 6
∴ 2ae = 6
∴ ae = `6/2` = 3
∴ a = `3/"e"` ...(i)
Distance between directrices = `(2"a")/"e"`
Given, distance between directrices = `50/3`
∴ `(2"a")/"e" = 50/3`
∴ `"a"/"e" = 25/3`
∴ `(3/"e")/"e" = 25/3` …[From (i)]
∴ `3/"e"^2 = 25/3`
∴ e2 = `9/25`
∴ e = `3/5` …[∴ 0 < e < 1]
Substituting e = `3/5` in (i), we get
a = `3/(3/5)`
∴ a = 5
∴ a2 = 25
Now, b2 = a2 (1 – e2)
= `25[1 - (3/5)^2]`
= `25(1 - 9/25)`
= `25(16/25)`
= 16
∴ The required equation of ellipse is `x^2/25 + y^2/16` = 1.
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संबंधित प्रश्न
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
Find the
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- length of the latus rectum
- distance between focii
- distance between directrices of the ellipse:
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- distance between directrices of the ellipse:
3x2 + 4y2 = 1
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