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Find the length of the latus rectum of ellipse whose foci are ( -2, -1) & (1, 2) and one of the directrices is x + y = 5

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Find the length of the latus rectum of ellipse whose foci are ( -2, -1) & (1, 2) and one of the directrices is x + y = 5
posted Nov 5, 2017 by anonymous

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1 Answer

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In an ellipse:
a = length of semi-major axis
b = length of semi-minor axis
c = distance from center to foci

Distance from center to directrix = a²/c
Length of latus rectum = 2b²/a

c = 1/2 distance from (−2,−1) to (1,2) = 1/2 √((1+2)²+(2+1)²) = 1/2 √18 = 3√2/2

Center = midpoint of foci = (−1/2, 1/2)
a²/c = Distance from (−1/2, 1/2) to line x + y − 5 = 0
a²/c = |−1/2 + 1/2 − 5| / √(1²+1²) = 5/√2
a² = 5/√2 c = (5/√2) (3√2/2) = 15/2

b² = a² − c² = 15/2 − 9/2 = 3

Length of latus rectum = 2b²/a = 2(3)/√(15/2) = 2√30/5

answer Feb 6 by Yasin Hossain Siinan



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