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