TS EAMCET 2017 Code A Solved Paper

since, x−y=−4k, or y=x+4k is a tangent to the parabola y2=8x, therefore
4k=‌‌

2

1

k=‌‌

1

2

Also point of contact p is (‌‌

a

m2

,‌‌

2a

m

)=(2,4)
Now the equation of normal to the y2=4ax at (x1,y1) is
y−y1=‌‌

−2

2a

(x−x1)
y−4=−1(x−2)
y−4=−x+2
x+y=6
x+y−6=0
The perpendicular distance of normal from (k,2k) that is
(‌‌

1

2

,1)
=‌‌

|‌‌ 1 2 +1−6|

√12+12

=‌‌

|‌‌ 3 2 −6|

√2

=‌‌

9

2√2