AP EAMCET Engineering 2012 Solved Paper

Given equation of parabola is
y2=12x
Here, a=3
∴ Equation of normal is
y=mx−2am−am3(∵m=−1)
y=−x−2(3)(−1)−3(−1)3
=−x+6+3
⇒ x+y=9 .......(i)
But given normal is
x+y=k
∴k=9
Focus of a given parabola is S(3,0).
Now, perpendicular distance from S(3,0) to the line (i) is
p=

|3(1)+0−9|

√12+12

=

6

√2

∴ 4k−2p2=4(9)−2(

6

√2

)2
=36−2×

36

2

=0