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

`((2)/(9), (8)/(9))`

Explanation

Let `A(t_(1)^(2), 2t_(1))` and `B (t_(2)^(2),2t_(2))` be coordnates of the end points of a chord of the parabola `y^(2) = 4x` having slope 2.
Now, slope fo AB is ,
`m = (2t_(2) - 2t_(1))/(t_(2)^(2) - t_(1)^(2)) = (2(t_(2) - t_(1)))/((t_(2) - t_(1))(t_(2) + t_(1))) = (2)/(t_(2) + t_(1))`
But m = 2
`implies 2 = (2)/(t_(2) + t_(1))` ,
`implies t_(1) + t_(2) = 1`
Let P (h,k) be a point on AB such that, it divides AB internally in the ratio 1 : 2.
Then, `h = (2t_(1)^(2) + t_(2)^(2))/(2 + 1)` and `k = (2(2t_(1)) + 2 t_(2))/(2 + 1)` ,
`implies 3h = 2t_(1)^(2) + t_(2)^(2)` .....(ii)
and `3k = 4t_(1) + 2t_(2)` .....(iii)
On substituting value of `t_(1)` from Eq. (i) Eq. (iii)
`3k = 4 (1 - t_(2)) + 2t_(2)`
`implies 3k = 4 - 2t_(2)`
`implies t_(2) = 2 - (3k)/(2)` .....(iv)
On substituting `t_(1) = 1 - t_(2)` in Eq. (ii) we get
`3h = 2 (1 - t_(2))^(2) + t_(2)^(2)`
`= 2 (1 - 2t_(2) + t_(2)^(2))+t_(2)^(2)`
`= 3t_(2)^(2) - 4t_(2) + 2 = 3 (t_(2)^(2) - (4)/(3) t_(2) + (2)/(3))`
`= 3 [(t_(2) - (2)/(3))^(2) + (2)/(3) - (4/(9)] = 3 (t_(2) - (2)/(3))^(2) + (2)/(3)`
`implies 3h - (2)/(3) = 3 (t_(2) - (2)/(3))^(2)`
`implies 3(h - (2)/(9)) = 3 (2 - (3k)/(2) - (2)/(3))^(2)` [from Eq. (iv)]
`implies 3 (h - (2)/(9)) = 3 ((4)/(3) - (3k)/(2))^(2)`
`implies (h - (2)/(9)) (9)/(4) (k - (8)/(9))^(2)`
`implies (k - (8)/(9))^(2) = (4)/(9) (h - (2)/(9))`
On generalising, we get the requried locus
`(y - (8)/(9))^(2) = (4)/(9) (x - (2)/(9))`
This represents a parabola with vertex at `((2)/(9),(8)/(9))`