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Coordinate Geometry

by Harmony
Tags: coordinate, geometry
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Harmony
#1
Aug3-07, 07:40 AM
P: 209
Question Statement
The straight line y=mx -2 intersects the curve y^2=4x at the two points P(x1,y1) and Q(x2,y2). Show that
1) m>-1/2, m not equal to 0.
2) x1 + x2 = 4(m+1)/m^2
3) y1 + y2 = 4/m

If the point O is the origin and the point T is a point such that OPTQ is a parallelogram, show that, when m changes, the equation of the locus of T is y^2 + 4y = 4x

My attempt:

The first part of the question is pretty straightforward, but the second part of the question (regarding the parallelogram) is a bit confusing.

My solution so far:
Let T be (x,y)
Gradient of OP = gradient of TQ

y1/x1 = y-y2 / x-x2
xy1 - x2y1 = yx1 - xy2
yx1 = xy1 + xy2 - x2y1......(1)

Similarly, Gradient of PT = gradient of OQ

yx2 = xy2 -xy2 + x2y1......(2)

(1) + (2)
(x1 + x2)y = (y1 + y2)x
4y(m+1)/ m^2 = 4x/m
y(m+1)/m = x

But I can't get rid of the m afterward
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#2
Aug3-07, 04:38 PM
P: 90
If your parellogram is in the form OPTQ this means simply T is the sum of the vectors P and Q
From this we conclude T=(x1+x2,y1+y2)
and from the first part you know the values
just try the equation given to you whether T is on that or not


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