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प्रश्न
Find the equation of the ellipse in standard form if eccentricity is `2/3` and passes through `(2, −5/3)`.
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उत्तर
Let the required equation of ellipse be `x^2/"a"^2 + y^2/"b"^2` = 1, where a > b.
Given, eccentricity (e) = `2/3`
The ellipse passes through `(2, -5/3)`.
∴ Substituting x = 2 and y = `-5/3` in equation of ellipse, we get
`2^2/"a"^2 + (-5/3)^2/"b"^2` = 1
∴ `4/"a"^2 + 25/(9"b"^2)` = 1
∴ `4/"a"^2 + 25/(9"a"^2(1 - "e"^2)` = 1 ...[∵ b2 = a2 (1 – e2)]
Multiplying throughout by a2, we get
`4 + 25/(9(1 - "e"^2)` = a2
∴ `4 + 25/(9[1 - (2/3)^2])` = a2
∴ `4 + 25/(9(1 - 4/9)` = a2
∴ `4 + 25/5` = a2
∴ 4 + 5 = a2
∴ a2 = 9
Now, b2 = a2 (1 – e2)
= `9[1 - (2/3)^2]`
= `9(1 - 4/9)`
= `9(5/9)`
= 5
∴ The required equation of ellipse is `x^2/9 + y^2/5` = 1.
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