हिंदी

Find the equation of the ellipse in standard form if eccentricity = 38 and distance between its foci = 6

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प्रश्न

Find the equation of the ellipse in standard form if eccentricity = `3/8` and distance between its foci = 6

योग
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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) = `3/8`

Distance between foci = 2ae

Given, distance between foci = 6

∴ 2ae = 6

∴ `2"a"(3/8)` = 6

∴ `(3"a")/4` = 6

∴ a = `(6 xx 4)/3` = 8

∴ a2 = 64

Now, b2 = a2 (1 – e2)

= `64[1 - (3/8)^2]`

= `64(1 - 9/64)`

= `64(55/64)`

= 55

∴ The required equation of the ellipse is `x^2/64 + y^2/55` = 1.

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अध्याय 7: Conic Sections - Exercise 7.2 [पृष्ठ १६३]

APPEARS IN

बालभारती Mathematics and Statistics (Arts and Science) Part 1 [English] Standard 11 Maharashtra State Board
अध्याय 7 Conic Sections
Exercise 7.2 | Q 2. (i) | पृष्ठ १६३
बालभारती Mathematics and Statistics (Arts and Science) Part 1 [English] Standard 11 Maharashtra State Board
अध्याय 7 Conic Sections
Miscellaneous Exercise 7 | Q II. (14) (i) | पृष्ठ १७८

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