Prove that the normal chord to a parabola at the point whose ordinate is equal to the abscissa subtends a right angle at the focus.

Explanation

Let the normal at P `(at_(1)^(2),2at_(1))` meet the curve at `Q(at_(2)^(2),2at_(2))`
`therefore` PQ is a normal chord and
`t_(2)=-t_(1)-2/(t_(1))`
By given condition `2at_(1)=at_(1)^(2)`
`thereforet_(1)=2`
from equation (i),
`t_(2)=-3`
then P(4a, 4a) and Q(9a, –6a) but focus S(a, 0)
`therefore` Slope of SP
`=(4a-0)/(4a-a)`
`=(4a)/(3a)=4/3`

and Slope of SQ
`=(-6a-0)/(9a-a)=(-6a)/(8a)`
`=-3/4`
`therefore` Slope of SP `x×` Slope of SQ
`=4/3xx-3/4=-1`
`thereforeanglePSQ=pi//2`
i.e. PQ subtends a right angle at the focus S.