(adsbygoogle = window.adsbygoogle || []).push({}); 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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# Coordinate Geometry

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