The curve be `y=x^2` whose slopeof tangent is x, so find the point of intersection of the tangent and the curve.

Answer

2

Explanation

2
`y^(2)=8x`
tangents at the end points of latus rectum meet the directrix on the x-axis at (-2,0).
`Delta_(1)="Area of "DeltaABP`
`=(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.
Tangent at point (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`.