JEE Advanced Model Paper 8 with solutions for online practice

The tangent at any point A{2secθ , tanθ} is given by

xsecθ

2

-

ytanθ

1

= 1
It meets the line x - 2y = 0
⇒

xsecθ

2

-

ytanθ

1

= 1 ⇒ x =

2

secθ−tanθ

⇒ Q = (

2

secθ−tanθ

,

1

secθ−tanθ

) ..... (i)
Also, the tangent meets the line x + 2y = 0 at R so
⇒

x

2

sec θ +

x

2

tan θ = 1 ⇒ x =

2

secθ+tanθ

⇒ R = (

2

secθ+tanθ

,

−1

secθ+tanθ

) ..... (ii)
Now, CQ . CR = √

22+12

(secθ−tanθ)2

√

22+11

(secθ+tanθ)

⇒ CQ . CR = 5