The focus of the parabola `y = 2x^2 + x` is

Options

A. `2:1`
B. `3:2`
C. `4:3`
D. `3:1`

Explanation

(4)
For `y^(2)=4x`, the coordinates of the ends of latus rectum are P(1,2) and Q(1,-2).

`DeltaPAQ` is isosceles right - angled. Therefore, the slope of PA is -1 and its equation is y-2=-(x-1), i.e., x+y-3=0.
Similarly, the equation of line QB is x-y-3=0.
Solving x+y-3=0 with the parabola `y^(2)=4x`, we have
`(3-x)^(2)=4x,i.e.,x^(2)-10x+9=0`
`:.x=1,9`
Therefore, the coordinates of B and C are (9,-6) and (9,6), respectively.
Area of trapezium `PBCQ=(1)/(2)xx(12+4)xx8=64` sq. units
Let the circumcenter of trapezium PBCQ be T(h,0). Then,
PT=BT
`orsqrt((h-1)^(2)+4)=sqrt((h-9)^(2)+36)`
`or-2h+=-18h+81+36`
`or16h=112`
`orh=7`
Hence, radius is `sqrt(40)=2sqrt(10`.
Let the inradius of `DeltaAPQ` be `r_(1)`. Then,
`r_(1)=(Delta_(1))/(s_(1))`
`((1)/(2)xx4xx2)/((1)/(2)(4+2sqrt(4+4)))`
`=(2)/(1+sqrt(2))=2(sqrt(2)-1)`
Let the inradius of `DeltaABC` be `r_(2)`. Then,
`r_(2)=(Delta_(2))/(s_(2))`
`=((1)/2xx12xx6)/((1)/(2)(12+2sqrt(36+36)))`
`=(6)/(1+sqrt(2))=6(sqrt(2)-1)`
`:." "(r_(2))/(r_(1))=3`