ABCD is a square with side AB = 2. A point P moves such that its distance from A equals its distance from the line BD. The locus of P meets the line AC at `T_1` and the line through A parallel to BD at `T_2` and `T_3`. The area of the triangle `T_1 T_2 T_3` is :

Options

A. `(1)/(2)`sq unit
B. `(2)/(3)`sq unit
C. 1 sq unit
D. 2sq units

Answer

C

Explanation

Since, `AG = sqrt2`
`AT_(1) = T_(1)G = (1)/(sqrt2)`
As, A is the focus, `T_(1)` is the vertex and BD I s the directrix of parabola.

Also, `T_(2)T_(3)` is latusrectum.
`therefore T_(2)T_(3) = 4 .(1)/(sqrt2)`
`therefore "Area of " DeltaT_(1)T_(2)T_(3)-(1)/(2)xx(1)/(sqrt2)xx(4)/(sqrt2)=1` sq unit