AP EAMCET Engineering 2011 Solved Paper

Let P(t2,2t) be the one end of a focal chord PQ of the parabola y2=4x, the coordinate of the other end Q are (

1

t2

,

−2

t

). ‌‌[∵tt′=−1]
∴ PQ=√(t2−

1

t2

)2+(2t+

2

t

)2
=(t+

1

t

)√(t−

1

t

)2+4
PQ=(t+

1

t

)2 .......(i)
Given, the chord makes a θ with positive direction of x -axis
⇒ tan‌θ=

−2∕t−2t

1∕t2−t2

=

−2

(1∕t−t)

=(t−1∕t)=2‌cot‌θ
Now from Eq. (i)
PQ=(t+1∕t)2
PQ=(t−1∕t)2+4
=(2‌cot‌θ)2+4
=4(cot2θ+1)
=4cosec2θ