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# Find the area of the region bounded by the latus recta of the ellipse x2/a2+y2/b2=1 and the tangents to the ellipse drawn at their ends.

Vikas TU
14149 Points
4 years ago
Dear Student,
Assuming hoz ellipse i.e a greater than b:
the tangents meet at y axis, so the region is a hexagon which can be divided symmetrically into 2 equal trapeziums.
distance between the 2 points of intersections of tangents on y-axis= a-(-a) =2a
distance from origin to focus= ae
end points of latus rectum = (ae, b^2/a) and (ae,-b^2/a)
length of latus rectum = 2b^2 /a
so area of a trapezium= ½ * (ae) * ((2b^2/a)+2a)
so total area = 2*½ * (ae) * ((2b^2/a)+2a)
=2e(a^2+b^2)
,where e is eccentricity= sqrt(1 – (b^2/a^2)) [Ans]
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)