The locus of the circumcenter of a variable triangle having sides the y-axis, y=2, and lx+my=1, where (1,m) lies on the parabola `y^(2)=4x`, is a curve C.
The curve C is symmetric about the line

Options

A. `x=3//2`
B. `y=-3//2`
C. `x=-3//2`
D. `y=3//2`

Answer

D

Explanation

(4)

`C-=(0,(1)/(m)),B-=((1-2m)/(l),2),A(0,2)`
Let (h,k) be the circumcenter of `DeltaABC` which is the midpoint of BC. Then,
`h=(1-2m)/(2l),k=(1+2m)/(2m)`
`orm=(1)/(2k-2),l=(k-2)/(2h(k-1))`
Given that (l,m) lies on `y^(2)=4x`. Then,
`m^(2)=4l`
`or((1)/(2k-2))^(2)=4{(k-2)/(2h(k-1))}`
`orh=8(k^(2)-3k+2)`
Therefore, the locus of (h,k) is
`x=8(y^(2)-3y+2)`
`or(y-(3)/(2))^(2)=(1)/(8)(x+2)`
Therefore, the vertex is `(-2,3//2)`.
Length of smallest focal chord =Length of latus rectum `=(1)/(8)`.
From the equation of curve C, it is clear that it is symmetric about the line y=3/2.