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Through the vertex O of a parabola y2=4x, chords OP & 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. Find the fixed point and the locus of the middle point of PQ.

agamya goyal , 9 Years ago

Grade 11

1 Answers

Anoopam Mishra

Last Activity: 9 Years ago

Let P be $(t^2_{1},2t_{1})$ and Q be $(t^2_{2},2t_{2})$ .

The slope of line OP is $2/t_{1}$ and that of OQ is $2/t_{2}$ .

Since they are at right angle then $2/t_{1} * 2/t_{2} = -1$ and hence $t_{1} * t_{2} = -4$ .

The coordinates of Q become $(16/t^2_{1}, -8/t_{1})$ .

The slope of the line PQ is $(2t_{1}+8/t_{1})/(t^2_{1}-16/t^2_{1}) = 2t_{1}/t^2_{1}-4$ .

The equation of line PQ is $y-2t_{1} = 2t_{1}/t^2_{1}-4 \ (x-t^2_{1})$ .

Substituting y=0, we get x=4.

Now the second part of the question, let the midpoint of PQ be (h,k).

Then, $2h = t^2_{1}+16/t^2_{1} \ and \ 2k = 2t_{1}-8/t_{1}$ .

$4k^2 = 4t^2_{1}+64/t^2_{1}-32 = 4(2h)-32$ .

=> $k^2 = 2(h-4)$

Hence the locus of the midpoint of PQ is $y^2 = 2(x-4)$ .

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