AP EAMCET Engineering 2013 Solved Paper

Given equations of straight lines are
x‌sec‌θ−y‌cosec‌θ=a ......(i)
x‌cos‌θ+y‌sin‌θ=a‌cos‌2‌θ
Also,
p = Perpendicular distance from the origin to the line (i)
P=

|0−0−a|

√sec2θ+cosec2θ

=

a‌sin‌θ.cos‌θ

√1

=a‌sin‌θ.cos‌θ=

a

2

‌sin‌2‌θ
⇒ 2p=a‌sin‌2‌θ
and q = perpendicular distance from the origin to the line (ii)
q=

|0+0−a‌cos‌2‌θ|

√cos2θ+sin2θ

=

a‌cos‌2‌θ

√1

=a‌cos‌2‌θ
Now,
4p2+q2=a2sin22θ+a2cos22θ
=a2(sin22θ+cos22θ)
=a2(1)=a2