EAMCET Engineering 2010 Solved Paper

Let (a secθ,b‌tan‌θ) be any point on the hyperbola ‌

x2

a2

−‌

y2

b2

=1
The equation of the asymptotes of the given hyperbola are
(‌

x

a

+‌

y

b

)=0‌ and ‌(‌

x

a

−‌

y

b

)=0
Now, P1= length of the perpendicular from (a secθ,b‌tan‌θ) on
⇒‌‌P1=‌

( x a + y b )=0

√‌ 1 a2 +‌ 1 b2

. . . (i)
P2= length of the perpendicular from (a secθ,b‌tan‌θ) on
‌‌(‌

x

a

−‌

y

b

)=0
⇒‌‌P2=‌

secθ−tan‌θ

√‌ 1 a2 +‌ 1 b2

. . . (ii)
∴‌‌P1P2=‌

( secθ+tan‌θ)

√‌ 1 a2 +‌ 1 b2

⋅‌

( secθ−tan‌θ)

√‌ 1 a2 +‌ 1 b2

=‌‌

( sec2θ−tan2θ)

(‌ 1 a2 +‌ 1 b2 )

=‌

1

(a2+b2) a2b2

⇒‌‌P1P2=‌

a2b2

a2+b2