Consider the parabola `y^(2) = 8x`. Let `Delta_(1)` be the area of the triangle formed by the end points of its latus rectum and the point `P(1/2,2)` on the parabola, and `Delta_(2)` be the area of the triangle formed by drawing tangents at P and at the end points of the latus rectum. Then `(Delta_(1))/(Delta_(2))` is

Answer

2

Explanation

`y^(2) = 8x `
Tangents at the endpoints of latus rectum meet the directrix on the -axis at (–2, 0)
`Delta_1 = "Area of "DeltaABQ = (1)/(2)(8)(2-(1)/(2))=6 , Delta_2 = " Area of "DeltaCQR`
Now , tangent at point `A(2,4) ` is ` 4y =4(x+2) or y =x +2 `
Also , tangent at point `P((1)/(2),2)` is `2y =4(x + (1)/(2)) " or " y =2x +1 `
Solving for Q and R , we get Q(1,3) and R(-1,-1)
Hence , area of `DeltaCQR " is " (1)/(2) ||{:(-1,-1,1),(1,3,1),(-2,0,1):}||=(1)/(2)(-3 + 2 + 6 + 1)=3`
Hence , the ratio of area is ` (6)/(3) =2 `.