If p,q are the perpendicular distances from the origin to the lines xsecα+ycosecα = a and xcosα - ysinα = acos2α respectively then show that 4p2 + q2 = a2
-
UP 0 DOWN 0 0 1
1 Answers
11p = -a/√(sec2α + cosec2α)
q = -aCos2α/√(cos2α+sin2α)
sec2α + cosec2α = 1/cos2α + 1/Sin2α = 1/Cos2αSin2α
p2 = a2Cos2αSin2α
q2 = a2Cos22α
4Cos2αSin2α = Sin22α
Now
=4p2 + q 2
=4a2Cos2αSin2α +a2Cos22α
= a2(Sin22α + Cos22α)
= a2
Hence proved