Show that the locus of a point that divides a chord of slope 2 of the parabola `y^2=4x` internally in the ratio `1:2` is parabola. Find the vertex of this parabola.

Answer

Vertex `-=((2)/(9),(8)/(9))`

Explanation

Slope of the chord joining point `P(t_(1))andQ(t_(2))` on the parabola `y^(2)=4x` is
`(2)/(t_(2)+t_(1))=2` (given)
`:." "t_(2)+t_(1)=1` (1)
Let point R(h,k) divide PQ in the ratio 1 : 2.
`So," "h=(t_(2)^(2)+2t_(1)^(2))/(1+2)andk=(2t_(2)+2(2t_(1)))/(1+2)`
`rArr" "3h=t_(2)^(2)+2t_(1)^(2)` (2)
`and" "t_(2)+2t_(1)=(3k)//2` (3)
From (1) and (3), we get
`t_(1)=(3)/(2)k-1andt_(2)=2-(3)/(2)k`
Putting these values in (2), we get
`(2-(3)/(2)k)^(2)+2((3)/(2)k-1)^(2)=3h`
Hence, locus of the point R is
`(y-(8)/(9))^(2)=(4)/(9)(x-(2)/(9))`,
which is parabola having vertex at `((2)/(9),(8)/(9))`.