`AB` is a chord of `y^2=4x` such that normals at `A` and `B` intersect at `C(9,6)`.

Options

A. `{:((p),(q),(r),(s).),(3,1,2,4):}`
B. `{:((p),(q),(r),(s).),(2,3,4,1):}`
C. `{:((p),(q),(r),(s).),(4,2,1,3):}`
D. `{:((p),(q),(r),(s).),(2,4,3,1):}`

Answer

A

Explanation

Normal at A and B intersect at `C(9,6)` which lies on parabola. Equation of normal is `y =- tx +2t +t^(3)`. Since it passes through C, we have
`t^(3) - 7t - 6 =0`
`:. T_(1) + t_(2) +3 =0` or `t_(1) + t_(2) =- 3`,
`t_(1)t_(2)t_(3) =6`, and `t_(1)t_(2) =2`.
Now, `AB = sqrt((t_(1)^(2)-t_(2)^(2))^(2)+(2t_(1)-2t_(2))^(2))`
`=sqrt((t_(1)-t_(2))^(2)[(t_(1)+t_(2))^(2)+4])`
`=sqrt([(t_(1)+t_(2))^(2)-4t_(1)t_(2)][(t_(1)+t_(2))^(2)+4])`
`=sqrt([9-8][9+4]) =sqrt(13)`
Area of triangle `ABC = (1)/(2) |{:(t_(1)^(2),2t_(1),1),(t_(2)^(2),2t_(2),1),(9,6,1):}|`
`= (1)/(2) [18(t_(1)-t_(2)) -6(t_(1)^(2)-t_(2)^(2))+2t_(1)t_(2)(t_(1)-t_(2)]`
`= (1)/(2) |t_(1)-t_(2)| [18 -6(t_(1)+t_(2))+2t_(1)t_(2)]`
= 20 sq. units
Equation of `AB: y - 2t_(2) = (2)/(t_(1)+t_(2)) (x-t_(2)^(2))`
Or `(t_(1)+t_(2))y -2t_(1)t_(2) - 2t_(2)^(2) = 2x - 2t_(2)^(2)` or `2x +3y +4 =0`
Distance of AB from origin is `(4)/(sqrt(13))`.
Line AB meets axis at `C(-2,0)` and `D(0,-4//3)`
`:.` Area of `DeltaOCD = 4//3` sq. units