Through the vertex 'O' of parabola `y^2=4x`, chords OP and OQ are drawn at right angles to one another. Show that for all positions of P, PQ cuts the axis of the parabola at a fixed point. Also find the locus of the middle point of PQ.

Answer

`y^(2) = 2 (x - 4)`

Explanation

Let the equation of chord OP be y = mx
Then, equation of chord will be `y = - (1)/(m)x` and P is point of interection of y - mx and `y^(2) = 4x` is
`((4)/(m^(2)), (4)/(m))` and Q is point intersection of `y = - (1)/(m) x` and `y^(2) = 4x` is `(4m^(2), -4m)`
Now, equation of PQ is
`y + 4m = ((4)/(m) + 4m)/((4)/(m^(2)) - 4m^(2)) (x - 4m^(2))`
`implies y + 4m = (m)/(1 - m^(2)) (x - 4m^(2))`
`implies (1 - m^(2)) y + 4m - 4m^(3) = mx - 4m^(3)`
`implies mx - (1 - m^(2)) y - 4m = 0`
This line meets X-axis where y = 0
i.e., `x = 4 implies OL= = 4` which is costant as indepdedent of m.
Again let (h,k) be the mid-point of PQ. Then
`h = (4m^(2) + (4)/(m^(2))/(2)`
and `k ((4)/(m) - 4m)/(2)`
`implies h = 2 (m^(2) + (1)/(m^(2)))`
and `k = 2 ((1)/(m) - m)`
`implies h = 2 [(m - (1)/(m))^(2) + 2]`
and `k = 2 ((1)/(m) - m)`
Eliminating m, we get
`2h = k^(2) + 8`
or `y^(2) = 2 (x - 4)` is required equation of locus