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1988 Paper 3 Q8
D: 1700.0 B: 1500.0
conics parabola tangent-normal parametric determinant area of triangle concurrent lines

Find the equations of the tangent and normal to the parabola \(y^{2}=4ax\) at the point \((at^{2},2at).\) For \(i=1,2,\) and 3, let \(P_{i}\) be the point \((at_{i}^{2},2at_{i}),\) where \(t_{1},t_{2}\) and \(t_{3}\) are all distinct. Let \(A_{1}\) be the area of the triangle formed by the tangents at \(P_{1},P_{2}\) and \(P_{3},\) and let \(A_{2}\) be the area of the triangle formed by the normals at \(P_{1},P_{2}\) and \(P_{3}.\) Using the fact that the area of the triangle with vertices at \((x_{1},y_{1}),(x_{2},y_{2})\) and \((x_{3},y_{3})\) is the absolute value of \[ \tfrac{1}{2}\det\begin{pmatrix}x_{1} & y_{1} & 1\\ x_{2} & y_{2} & 1\\ x_{3} & y_{3} & 1 \end{pmatrix}, \] show that \(A_{3}=(t_{1}+t_{2}+t_{3})^{2}A_{1}.\) Deduce a necessary and sufficient condition in terms of \(t_{1},t_{2}\) and \(t_{3}\) for the normals at \(P_{1},P_{2}\) and \(P_{3}\) to be concurrent.

Solution: \(\frac{dy}{dt} = 2a, \frac{dx}{dt} = 2at \Rightarrow \frac{dy}{dx} = \frac{1}{t}\). Therefore the equation of the tangent will be \(\frac{y - 2at}{x-at^2} = \frac{1}{t} \Rightarrow y = \frac1tx +at\) and normal will be \(\frac{y-2at}{x-at^2} = -t \Rightarrow y = t(at^2-x+2a)\). The tangents will meet when: \begin{align*} && \begin{cases} t_iy -x &= at_i^2 \\ t_j y - x &= at_j^2 \\ \end{cases} \\ \Rightarrow &&(t_i - t_j)y &= a(t_i-t_j)(t_i+t_j) \\ \Rightarrow && y &= a(t_i+t_j) \\ && x &= at_it_j \end{align*} The normals will meet when: \begin{align*} && \begin{cases} y +t_i x &= at_i^3+2at_i \\ y +t_j x &= at_j^3+2at_j \\ \end{cases} \\ \Rightarrow &&(t_i - t_j)x &= a(t_i-t_j)(t_i^2+t_it_j+t_j^2+2) \\ \Rightarrow && x&= a(t_i^2+t_it_j+t_j^2+2) \\ && y &= -at_it_j(t_i+t_j) \end{align*} Therefore the area of our triangles will be: \begin{align*} \tfrac{1}{2}\det\begin{pmatrix}at_1t_2 & a(t_1+t_2) & 1\\ at_2t_3 & a(t_2+t_3) & 1\\ at_3t_1 & a(t_3+t_1) & 1 \end{pmatrix} &= \frac{a^2}{2}\det\begin{pmatrix}t_1t_2 & (t_1+t_2) & 1\\ t_2t_3 & (t_2+t_3) & 1\\ t_3t_1 & (t_3+t_1) & 1 \end{pmatrix} \\ &= \frac{a^2}{2}\det\begin{pmatrix}t_1t_2 & (t_1+t_2) & 1\\ t_2(t_3-t_1) & (t_3-t_1) & 0\\ t_1(t_3-t_2) & (t_3-t_2) & 0 \end{pmatrix} \\ &= \frac{a^2}{2}|(t_2-t_1)(t_3-t_2)(t_1-t_3)| \end{align*} and \begin{align*} \tfrac{1}{2}\det\begin{pmatrix}a(t_1^2+t_1t_2+t_2^2+2) & -at_1t_2(t_1+t_2) & 1\\ a(t_2^2+t_2t_3+t_3^2+2) & -at_2t_3(t_2+t_3) & 1\\ a(t_3^2+t_3t_1+t_1^2+2) & -at_3t_1(t_3+t_1) & 1\\ \end{pmatrix} &= \frac{a^2}{2}\det\begin{pmatrix}(t_1^2+t_1t_2+t_2^2+2) & -t_1t_2(t_1+t_2) & 1\\ (t_2^2+t_2t_3+t_3^2+2) & -t_2t_3(t_2+t_3) & 1\\ (t_3^2+t_3t_1+t_1^2+2) & -t_3t_1(t_3+t_1) & 1\\ \end{pmatrix} \\ &= \frac{a^2}{2}\det\begin{pmatrix}(t_1^2+t_1t_2+t_2^2+2) & -t_1t_2(t_1+t_2) & 1\\ t_3^2-t_1^2+t_2(t_3-t_1) & t_2(t_1^2+t_1t_2-t_2t_3-t_3^2) & 0\\ t_3^2-t_2^2+t_1(t_3-t_2) & t_1(t_2^2+t_2t_1-t_1t_3-t_3^2) & 0\\ \end{pmatrix} \\ &= \frac{a^2}{2}\det\begin{pmatrix}(t_1^2+t_1t_2+t_2^2+2) & -t_1t_2(t_1+t_2) & 1\\ (t_3-t_1)(t_3+t_2+t_1) & t_2(t_1-t_3)(t_1+t_3+t_2) & 0\\ (t_3-t_2)(t_3+t_2+t_1) & t_1(t_2-t_3)(t_1+t_2+t_3)& 0\\ \end{pmatrix} \\ &= \frac{a^2}{2}(t_1+t_2+t_3)^2|(t_2-t_1)(t_3-t_2)(t_1-t_3)| \end{align*} as required. The normals will be concurrent iff the area of their triangle is \(0\). This is certainly true if \(t_1+t_2+t_3 = 0\). In fact the only if is also true, since no \(3\) tangents can be concurrent.

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1988 Paper 3 Q14
D: 1700.0 B: 1484.0
circular motion bead on wire normal reaction radius of curvature parabola energy conservation projectile motion constraint motion

A small heavy bead can slide smoothly in a vertical plane on a fixed wire with equation \[ y=x-\frac{x^{2}}{4a}, \] where the \(y\)-axis points vertically upwards and \(a\) is a positive constant. The bead is projected from the origin with initial speed \(V\) along the wire.

  1. Show that for a suitable value of \(V\), to be determined, a motion is possible throughout which the bead exerts no pressure on the wire.
  2. Show that \(\theta,\) the angle between the particle's velocity at time \(t\) and the \(x\)-axis, satisfies \[ \frac{4a^{2}\dot{\theta}^{2}}{\cos^{6}\theta}+2ga(1-\tan^{2}\theta)=V^{2}. \]

Solution:

  1. The condition that the bead exerts no pressure on the wire is equivalent to the condition that the wire exerts no force on the bead. (Newton's Third Law). This is equivalent to the bead being projected under gravity. Notice that the initial projection is at \(45^{\circ}\) since \(\frac{dy}{dx}|_{x=0} = 1\). The position of the particle (under gravity) at time \(t\) is \(x = \frac{1}{\sqrt{2}}Vt\) and \(y = \frac{1}{\sqrt{2}}Vt - \frac12 gt^2 = x - \frac{1}{2}g \frac{2x^2}{V^2} = x - \frac{g}{V^2}x^2\). Therefore they follow the same trajectory if \(\frac{g}{V^2} = \frac{1}{4a} \Leftrightarrow V = 2\sqrt{ag}\)
  2. First note that the wire does no work as it is perpendicular to the velocity, so it is fine to use conservation of momentum. If we take our \(0\) GPE level to be be \(x = 0\), then we notice the initial energy is \(\frac12mV^2\). Secondly, notice that \(\tan \theta = \frac{\d y}{\d x} = 1- \frac{x}{2a} \Rightarrow x = 2a - 2a \tan \theta\) \begin{align*} y &= 2a(1-\tan \theta) - \frac{4a^2(1-\tan \theta)^2}{4a}\\ &= (1-\tan \theta)(2a-a(1-\tan \theta)) \\ &= a(1-\tan \theta)(1+\tan \theta) \\ &= a(1-\tan^2 \theta) \end{align*} GPE \(mga(1-\tan^2 \theta)\). To calculate the kinetic energy, notice that \(\dot{x} = v \cos \theta \dot{\theta}\) and \(\dot{x} = -2a\sec^2 \theta\dot{\theta} \Rightarrow v = -\frac{2a\dot{\theta} }{\cos^{3} \theta}\). Therefore, energy at time \(t\) is: \begin{align*} && \frac12 m V^2 &= \frac12 m \l - \frac{2a\dot{\theta}}{\cos^3 \theta} \r^2 + mga(1-\tan^2 \theta) \\ \Rightarrow && V^2 &= \frac{4a^2\dot{\theta}^2}{\cos^6 \theta} + 2ag(1-\tan^2 \theta) \end{align*}

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