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Hi greenlemings, I confess to being a newby both to this forum and parametric equations, however my study to date leads me to offer the following solutions and also to refer you to this document which I have found invaluable in my learning...

I would really appreciate it if someone could review my answers for 1c) and 2c). I think i have taken the correct approaches but would like to be sure. 1 c) (i) Convert R(a(p+q),apq) to Cartesian where pq = -2 Sub for pq in y = apq => y = -2a (ii) Convert R(a(a+p),apq) to...

Thanks Integrand for the quick response. I thought that might be the case as in 1ci) and 1ciii) I could not get real values when solving for p & q simultaneously between (i) & (iii). But I thought I would seek expert opinion before continuing. I will now return to the problem and see how I go...

Hi, I had been progressing well with my understanding of parametrics until these two questions arose. The questions in full are: 1) Tangents to the parabola x^2 = 4ay at the points P(2ap,ap^2)and Q(2aq,aq^2) meet at R. a) Show that the equation of the tangent at P is y - px + ap^2 = 0...

Ah, the wonders of the mighty computer over the visualisation of my mind. You are indeed correct and I have been chasing a phantom all this time. I had the correct answer all the time. Looks like my time would have been better spent learning to use GeoGebra. Many thanks for your graphical...

My question arises from the following: The normal at any point P (2at,at^2) on the parabola x^2 = 4ay cuts the y -xis a Q and is projected to R so that PQ=QR A) Find the coordinates of R in terms of t. My answer at this point is R(-2at,at^2 + 4a) ) I arrived at that by identifying the...

InteGrand, Therein lies the problem. The parabola thus defined is open at the top and I know the locus of the point is on an inverted parabola as I have physically drawn a sketch to confirm the locus. Graeme

Hi, I am working on a problem where I know a point (-2at,at^2 +4a) lies on an inverted parabola about the y-axis. Normally I would transform x = -2at to t = -x/2a and then sub for t in y = at^2 + 4a to get to the Cartesian form. But this does not work in this instance. I think it has...