BITSAT 2014 Paper

We have P1 = length of perpendicular from (0, 0) on x secθ + y cosecθ = a
i.e. P1 = |

a

√sec2θ+cosec2θ

| = |a sinθ cosθ|
= |

a

2

sin2θ| or 2 P1 = |a sin 2 θ|
P2= Length of the perpendicular from (0, 0) on
x cos θ - y sin θ = a cos 2θ
P2 = |

acos2θ

√cos2θ+sinθ

| = |a cos 2θ|
Now, 4P12+P−22 = a2sin22θ+a2cos22θ = a2