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Find the equation of the ellipse in standard form if eccentricity is 23 and passes through (2,−53). - Mathematics and Statistics

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Question

Find the equation of the ellipse in standard form if eccentricity is `2/3` and passes through `(2, −5/3)`.

Sum
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Solution

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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Conic Sections - Ellipse
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Chapter 7: Conic Sections - Exercise 7.2 [Page 163]

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