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Question

Show that the product of perpendiculars on the line xa cos θ+yb sin θ=1 from the points (±a2-b2,0) is b2.

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Solution

Let d1 and d2 be the perpendicular distances of line xa cos θ+yb sin θ=1 from points a2-b2,0 and -a2-b2,0, respectively.



d1=a2-b2acosθ-1cos2θa2+sin2θb2=b a2-b2cosθ-aa2sin2θ+b2cos2θ

Similarly,

d1=-a2-b2acosθ-1cos2θa2+sin2θb2=b -a2-b2cosθ-aa2sin2θ+b2cos2θ=b a2-b2cosθ+aa2sin2θ+b2cos2θ

Now,

d1d2=b a2-b2cosθ-aa2sin2θ+b2cos2θ×b a2-b2cosθ+aa2sin2θ+b2cos2θd1d2=b2 a2-b2cos2θ-a2a2sin2θ+b2cos2θd1d2=b2 a2cos2θ-1-b2cos2θa2sin2θ+b2cos2θd1d2=b2-a2sin2θ-b2cos2θa2sin2θ+b2cos2θ=b2a2sin2θ+b2cos2θa2sin2θ+b2cos2θ=b2

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