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Question
If p1 and p2 are the lengths of the perpendiculars from the origin to the straight lines x sec θ + y cosec θ = 2a and x cos θ – y sin θ = a cos 2θ, then prove that p12 + p22 = a2
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Solution
Given P1 is the length of the perpendicular from the origin to the straight line
x sec θ + y cosec θ – 2a = 0
P1 = `(0 * sec theta + 0 * "cosec" theta - 2"a")/sqrt(sec^2theta + "cosec"^2theta)`
P12 = `(4"a"^2)/(sec^2theta + "cosec"^2theta)` .....(1)
Also given P2 is the length of the perpendicular from the origin to the straight line
x cos θ – y sin θ – a cos 2θ = 0
P2 = `(0 * cos theta - 0 sin theta - "a" cos 2theta)/sqrt(cos^2theta + (- sin theta)^2`
P2 = `(- "a" cos 2theta)/sqrt(cos^2theta + sin^2theta)`
= – a cos 2θ
P22 = a2 cos2 2θ ......(2)
P12 + P22 = `(4"a"^2)/(sec^2theta + "cosec"^2theta) + "a"^2cos^2 2theta`
= `(4"a"^2)/(1/(cos^2theta) + 1/(sin^2theta)) + "a"^2(cos^2theta - sin^2theta)`
= `(4"a"^2)/((sin^2theta + cos^2theta)/(cos^2theta * sin^2theta)) + "a"^2(cos^4theta + sin^4theta - 2cos^2theta sin^2theta)`
= 4a2 cos2θ sin2θ + a2cos4θ + a2sin4θ - 2a2cos2θ sin2θ
= a2cos4θ + a2sin4θ + 2a2 cos2θ sin2θ
= a2[cos4θ + sin4θ + 2cos2θ sin2θ]
= a2[cos2θ + sin2θ]2
P12 + P22 = a2
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