AP EAMCET Engineering 2013 Solved Paper

Given, equation of parabola is
y2=8x ......(i)
∴ a=2
Let (h,4) be the coordinate of mid-point of chord. Then, equation of chord is
y−4=m(x−h) ......(ii)
If line (ii) passes through the point P(2t12,4t1) and Q(2t22,4t2) on parabola Eq. (i), then
y(t1+t2)−2x−4t1t2=0 .......(iii)
having slope
m=

2

t1+t2

......(iv)
Since, (h,4) is the mid-point of PQ. Therefore,
2×4=4(t1+t2)
⇒ t1+t2=2
Hence, slope of chord PQ ism=

2

2

=1 [using Eq. (iv)]