Correct Answer - A
Let R be the point of intersection of tangents at P and Q. Then, the coordinates of R are `(at_(1), t_(2), a(t_(1)+t_(2)))`.
`:." Area of "Delta PQR=1/2|Delta|`, where
`Delta=|{:(at_(1)^(2),2at_(1), 1),(at_(1)^(2),2at_(2),1),(at_(1)t_(2),a(t_(1)+t_(2)),1):}|`
`rArrDelta=a^(2)|{:(t_(1)^(2),2t_(1),1),(t_(2)^(2),2t_(2),1),(t_(1)t_(2),t_(1)+t_(2),1):}|`
`rArrDelta=a^(2)|{:(t_(1)^(2)-t_(1)t_(2),t_(1)-t_(2),0),(t_(2)^(2)-t_(1)t_(1),t_(2)-t_(1),0),(t_(1)t_(2),t_(1)+t_(2),1):}|[{:("Appliyin",R_(1)toR_(1)-R_(3)),(,R_(2)toR_(2)-R_(3)):}]`
`rArrDelta=a^(2)|{:(t_(1)(t_(1)-t_(2)),t_(1)-t_(2),0),(-t_(2)(t_(1)-t_(2)),-(t_(1)-t_(2)),0),(t_(1)t_(2),t_(1)+t_(2),1):}|`
`rArrDelta=a^(2)(t_(1)-t_(2))^(2)|{:(t_(1),1,0),(-t_(2),-1,0),(t_(1)t_(2),t_(1)+t_(2),1):}|`
`rArrDelta=a^(2)(t_(1)-t_(2))^(2)(-t_(1)+t_(2))=-a^(2)(t_(1)-t_(2))^(3)`
Hence, area of `Delta PQR=1/2a^(2)|t_(1)-t_(2)|^(3)`