The given equation of line is

x

a

cos θ +

y

b

sin θ = 1 ... (i)
Now distance of (i) from the point (√a2−b2,0)
=

| √a2−b2 a cosθ−1|

√( cosθ a )2+( sinθ b )2

And distance of (i) from the point (−√a2−b2,0)
=

|− √a2−b2 a cosθ−1|

√( cosθ a )2+( sinθ b )2

Now, product of lengths of these two perpendiculars
=
=

|( a2−b2 a2 )cos2θ−1|

b2cos2θ+a2cos2θ a2b2

=

|(a2cos2θ−b2cos2θ)−a2|×a2b2

(b2cos2θ+a2sin2θ)a2

=

|a2(cos2θ−1)−b2cos2θ|×b2

a2sin2θ+b2cos2θ

=

|−a2sin2θ−b2cos2θ|b2

a2sin2θ+b2cos2θ

= |

−(a2sin2θ+b2cos2θ)

(a2sin2θ+b2cos2θ)

|b2 = |−1|b2 = b2
Hence proved.