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Coordinate Geometry 
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#1
Aug307, 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 = yy2 / xx2 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 


#2
Aug307, 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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