Which paper was that? I can imagine a sizable number of 3U candidates doing IBP as they might have learnt it in 4U but how many 2U students have come across it (unless they were accelerated and planned to do 4U)?
I would assume accelerants.
One of my students is in Year 11, and has already learnt IBP.
Like 2007 or something. It was the last question (Q10) that asked to find the area underneath a log graph.
<a href="http://www.codecogs.com/eqnedit.php?latex=\\\textrm{Equation of the tangent:}~y=tx-at^2\\~\\ \textrm{Finding points of intersection:}\\ x^2=-4a(tx-at^2)\\ x^2@plus;4atx-4a^2t^2=0\\ \begin{align*} x&=\frac{-4at \pm \sqrt{(4at)^2-4(-4a^2t^2)}}{2}\\ &=\frac{-4at \pm 4\sqrt{2}at}{2}\\ &=2at(-1 \pm 2\sqrt{2}) \end{align*}\\ \therefore \textrm{the x-coordinate of the midpoint of PQ is:}\\ \frac{2at(-1@plus;\sqrt{2})@plus;2at(-1-\sqrt{2})}{2}=-2at\\ ~\\ y=\frac{x^2}{4a}=\frac{4a^2t^2(-1 \pm \sqrt{2})^2}{4a}=at^2(-1 \pm \sqrt{2})^2\\ ~\\ \therefore\textrm{the y-coordinate of the midpoint of PQ is:}\\ \frac{at^2(-1@plus;\sqrt{2})^2@plus;at^2(-1-\sqrt{2})^2}{2}=3at^2\\ ~\\ t=\frac{x}{-2a}\\ \therefore y=3a(\frac{x}{-2a})^2=\frac{3x^2}{4a}\\ x^2=\frac{4a}{3}y ~\\ \textrm{The locus of the midpoint of PQ is a parabola with vertex (0,0) and focal length}~\frac{a}{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\\textrm{Equation of the tangent:}~y=tx-at^2\\~\\ \textrm{Finding points of intersection:}\\ x^2=-4a(tx-at^2)\\ x^2+4atx-4a^2t^2=0\\ \begin{align*} x&=\frac{-4at \pm \sqrt{(4at)^2-4(-4a^2t^2)}}{2}\\ &=\frac{-4at \pm 4\sqrt{2}at}{2}\\ &=2at(-1 \pm 2\sqrt{2}) \end{align*}\\ \therefore \textrm{the x-coordinate of the midpoint of PQ is:}\\ \frac{2at(-1+\sqrt{2})+2at(-1-\sqrt{2})}{2}=-2at\\ ~\\ y=\frac{x^2}{4a}=\frac{4a^2t^2(-1 \pm \sqrt{2})^2}{4a}=at^2(-1 \pm \sqrt{2})^2\\ ~\\ \therefore\textrm{the y-coordinate of the midpoint of PQ is:}\\ \frac{at^2(-1+\sqrt{2})^2+at^2(-1-\sqrt{2})^2}{2}=3at^2\\ ~\\ t=\frac{x}{-2a}\\ \therefore y=3a(\frac{x}{-2a})^2=\frac{3x^2}{4a}\\ x^2=\frac{4a}{3}y ~\\ \textrm{The locus of the midpoint of PQ is a parabola with vertex (0,0) and focal length}~\frac{a}{3}" title="\\\textrm{Equation of the tangent:}~y=tx-at^2\\~\\ \textrm{Finding points of intersection:}\\ x^2=-4a(tx-at^2)\\ x^2+4atx-4a^2t^2=0\\ \begin{align*} x&=\frac{-4at \pm \sqrt{(4at)^2-4(-4a^2t^2)}}{2}\\ &=\frac{-4at \pm 4\sqrt{2}at}{2}\\ &=2at(-1 \pm 2\sqrt{2}) \end{align*}\\ \therefore \textrm{the x-coordinate of the midpoint of PQ is:}\\ \frac{2at(-1+\sqrt{2})+2at(-1-\sqrt{2})}{2}=-2at\\ ~\\ y=\frac{x^2}{4a}=\frac{4a^2t^2(-1 \pm \sqrt{2})^2}{4a}=at^2(-1 \pm \sqrt{2})^2\\ ~\\ \therefore\textrm{the y-coordinate of the midpoint of PQ is:}\\ \frac{at^2(-1+\sqrt{2})^2+at^2(-1-\sqrt{2})^2}{2}=3at^2\\ ~\\ t=\frac{x}{-2a}\\ \therefore y=3a(\frac{x}{-2a})^2=\frac{3x^2}{4a}\\ x^2=\frac{4a}{3}y ~\\ \textrm{The locus of the midpoint of PQ is a parabola with vertex (0,0) and focal length}~\frac{a}{3}" /></a> $ meets the parabola $ x^{2}=-4ay $ at two points P and Q. Find the locus of the midpoint of PQ.$ )
Once you find the midpoint of the interval, all you need is the y value of the same interval, which can easily be found by substituting that x coordinate -2at into the equation of the tangent (since the midpoint of PQ lies on the tangent).
