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 November 30th, 2009, 05:18 AM #1 Senior Member   Joined: Sep 2008 Posts: 199 Thanks: 0 coordinate geometry Two variable points P(ap^2,2ap) and Q(aq^2,2aq) on the parabola y^2=4ax are such that PQ subtends a right angle at the origin. (1) Show that the chord PQ passes through a fixed point on the x-axis. Gradient of PQ = 2/(p+q) Equation of PQ: $y-2ap=\frac{2}{p+q}(x-ap^2)$ , when y=0, I found x=-apq, which means the chord PQ cuts the x-axis at (-apq,0) Am I correct? (2) Find the locus of the point of intersection of the tangents to the parabola at P and Q . I found the point of intersection of the tangents to be (apq , a(p+q)) and here I am not sure how to move on. THanks for helping.
November 30th, 2009, 11:41 PM   #2
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Re: coordinate geometry

Quote:
 Two variable points P(ap^2,2ap) and Q(aq^2,2aq) on the parabola y^2=4ax are such that PQ subtends a right angle at the origin.
That is, the position vectors OP and OQ are perpendicular. Take the dot product, equate it with zero, and show that pq is constant.

December 1st, 2009, 08:52 AM   #3
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Quote:
 Originally Posted by mikeportnoy I found x=-apq, which means the chord PQ cuts the x-axis at (-apq,0) . . . Am I correct?
Yes. Since OP?OQ, (2/p)(2/q) = -1, and so -pq = 4. Hence PQ passes through the fixed point (4a, 0).

Quote:
 Originally Posted by mikeportnoy I found the point of intersection of the tangents to be (apq , a(p+q))
Correct. Since -pq = 4, the x-coordinate is -4a. You can show the y-coordinate can have any value, so the required locus is the line x = -4a.

December 1st, 2009, 09:45 PM   #4
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Re:

Quote:
Originally Posted by skipjack
Quote:
 Originally Posted by mikeportnoy I found x=-apq, which means the chord PQ cuts the x-axis at (-apq,0) . . . Am I correct?
Yes. Since OP?OQ, (2/p)(2/q) = -1, and so -pq = 4. Hence PQ passes through the fixed point (4a, 0).

Quote:
 Originally Posted by mikeportnoy I found the point of intersection of the tangents to be (apq , a(p+q))
Correct. Since -pq = 4, the x-coordinate is -4a. You can show the y-coordinate can have any value, so the required locus is the line x = -4a.
THanks !

so x=apq , y=a(p+q) and the locus here would be x=-4a

I understood all this but I don't understand what u mean by 'You can show the y-coordinate can have any value'

 December 2nd, 2009, 12:58 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,370 Thanks: 2007 Please don't quote the entire post you are replying to if your reply would make just as good sense without your doing so, which is usually the case when replying to the immediate preceding post. If you want a(p + q) to have the value r, p and q must be the roots of t² - (r/a)t - 4 = 0. Whatever real value r has, that equation has real roots.

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