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Question

Prove that the product of the lengths of the perpendiculars drawn from the points (a2b2,0) and(a2b2,0) to the line xacosθ+ybsinθ=1 is b2.

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Solution

Perpendicular Length from Point to line P(x1,y1)
ax+by+c=0
d=|ax1+by1+c|a2+b2
given line xacosθ+ybsinθ=1
given Point (a2b2,0) and (a2b2,0)
Product of Perpendicular distance from
both Point to line
=a2b2acosθ+01cos2θa2+sin2θb2×a2b2acosθ+01cos2θa2+sin2b2
=cos2θa2(a2b2)+1cos2θa2+sin2θb2=(cos2θ+b2a2cos2θ+1)a2b2(b2cos2θ+a2sin2θ)
=(a2sin2θ+b2cos2θ)a2b2a2(a2sin2θ+b2cos2θ)=b2

1169919_875587_ans_5fe32498b8b943a0b84b6fdb94abce2c.jpg

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