`A B` is a double ordinate of the parabola `y^2=4a xdot` Tangents drawn to the parabola at `Aa n dB` meet the y-axis at `A_1a n dB_1` , respectively. If the area of trapezium `AA_1B_1B` is equal to `12 a^2,` then the angle subtended by `A_1B_1` at the focus of the parabola is equal to `2tan^(-1)(3)` (b) `tan^(-1)(3)` `2tan^(-1)(2)` (d) `tan^(-1)(2)`

Options

A. `2tan^(-1)` (3)
B. `tan^(-1)` (3)
C. `2tan^(-1)` (2)
D. `tan^(-1)` (2)

Answer

C

Explanation

(3) Let `A-=(at_(1)^(2),2at_(1))`
`andB-=(at_(1)^(2),-2at_(1))`
The equation of tangents at A and B are, respectively,
`t_(1)y=x+at_(1)^(2)`
`and-t_(1)y=x+at_(1)^(2)`
These tangents meet the y-axis, respectively, at
`A_(1)-=(0,at_(1))`
`andB_(1)-=(0,-at_(1))`
Area of trapezium
`A A_(1)B_(1)B=(1)/(2)(AB+A_(1)B_(1))xxOC`
`:." "24a^(2)=(1)/(2)(4at_(1)+2at_(1))(at_(1)^(2))`
`or" "t_(1)^(3)=8ort_(1)=2`
`:." "A_(1)-=(0,2a)`
If `angleOSA_(1)=theta`,then
`tantheta=(2a)/(a)=2`
Thus, the middle required angle is `2tan^(-1)2`.