The sum of the squares of the perpendiculars on any tangent to the ellipse x2/a2+y2/b2=1 from two points on the minor axis each at a distance a2−b2 from the centre is
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a
2a2
b
2b2
c
a2+b2
d
a2−b2
answer is A.
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Detailed Solution
The eccentricity e of the given ellipse is given by g2=1−b2/a2⇒a2−b2=a2e2 . So the point on the minoraxis, i.e. y-axis at a distance a2−b2 from the centre (0, 0) of the ellipse are (0, ± ae). The equation of the tangent at any point (a cos θ, b sin θ) on the ellipse is xacosθ+ybsinθ=1 So the required sum is aesinθb−1cos2θa2+sin2θb22+−aesinθb−1cos2θa2+sin2θb22=(aesinθ−b)2+(aesinθ+b)2b2cos2θ+a2sin2θ×a2=2a2a2e2sin2θ+b2b2cos2θ+a2sin2θ =2a2a2−b2sin2θ+b2b2cos2θ+a2sin2θ=2a2
The sum of the squares of the perpendiculars on any tangent to the ellipse x2/a2+y2/b2=1 from two points on the minor axis each at a distance a2−b2 from the centre is