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2014 Paper 1 Q8
Let \(L_a\) denote the line joining the points \((a,0)\) and \((0, 1-a)\), where \(0< a < 1\). The line \(L_b\) is defined similarly.
- Determine the point of intersection of \(L_a\) and \(L_b\), where \(a\ne b\).
- Show that this point of intersection, in the limit as \(b\to a\), lies on the curve \(C\) given by \[ y=(1-\sqrt x)^2\, \ \ \ \ (0< x < 1)\,. \]
- Show that every tangent to \(C\) is of the form \(L_a\) for some \(a\).
Solution:
- \(L_a : \frac{y}{x-a} = \frac{1-a-0}{0-a} = \frac{a-1}{a} \Rightarrow ay+(1-a)x = a(1-a)\) \begin{align*} && ay + (1-a)x &= a(1-a) \\ && by + (1-b)x &= b(1-b) \\ \Rightarrow && aby + b(1-a)x &= ba(1-a) \\ && aby + a(1-b)x &= ab(1-b) \\ \Rightarrow && (b-a)x &= ab(b-a) \\ \Rightarrow && x &= ab \\ && y &= \frac{a-1}{a} \cdot a(b-1) \\ &&&= (1-a)(1-b) \end{align*}
- As \(a \to b\), \(x \to a^2, y \to 1-2a+a^2 =(1-a)^2 = (1-\sqrt{x})^2\)
- \(\frac{\d y}{\d x} = 2(1-\sqrt{x})\cdot \left (-\tfrac12 \frac{1}{\sqrt{x}} \right) = 1 - \frac{1}{\sqrt{x}}\). Therefore the tangent when \(x = c^2, y = (1-c)^2\) is \begin{align*} && \frac{y-(1-c)^2}{x-c^2} &= 1 - \frac{1}{c} \\ \Rightarrow && cy + (1-c)x &= c(c-1)+c(1-c)^2 \\ &&&= c(1-c) \end{align*} Which is an equation of the form \(L_c\)
2013 Paper 2 Q4
The line passing through the point \((a,0)\) with gradient \(b\) intersects the circle of unit radius centred at the origin at \(P\) and \(Q\), and \(M\) is the midpoint of the chord \(PQ\). Find the coordinates of \(M\) in terms of \(a\) and \(b\).
- Suppose \(b\) is fixed and positive. As \(a\) varies, \(M\) traces out a curve (the locus of \(M\)). Show that \(x=- by\) on this curve. Given that \(a\) varies with \(-1\le a \le 1\), show that the locus is a line segment of length \(2b/(1+b^2)^\frac12\). Give a sketch showing the locus and the unit circle.
- Find the locus of \(M\) in the following cases, giving in each case its cartesian equation, describing it geometrically and sketching it in relation to the unit circle:
- \(a\) is fixed with \(0 < a < 1\), and \(b\) varies with \(-\infty < b < \infty\);
- \(ab=1\), and \(b\) varies with \(0< b\le1\).
Solution: \begin{align*} && y &= bx-ba \\ && 1 &= x^2 + y^2 \\ \Rightarrow && 1 &= x^2 + b^2(x-a)^2 \\ \Rightarrow && 0 &= (1+b^2)x^2-2ab^2x+b^2a^2-1 \end{align*} This will have roots which sum to \(\frac{2ab^2}{1+b^2}\), therefore \(M = \left ( \frac{ab^2}{1+b^2}, \frac{ab^3}{1+b^2}-ba \right)=\left ( \frac{ab^2}{1+b^2}, \frac{-ba}{1+b^2} \right)\)
- Since \(b\) is fixed so is \(\frac{b}{1+b^2} = t\) and all the points are \((bta, -ta)\), ie \(x = -by\). If \(a \in [-1,1]\) we are ranging on the points \((bt, -t)\) to \((-bt, t)\) which is a distance of \begin{align*}
&& d &= \sqrt{(bt+bt)^2+(-2t)^2} \\
&&&= \sqrt{4(b^2+1)t^2} \\
&&&=2 \sqrt{(b^2+1)\frac{b^2}{(b^2+1)^2}} \\
&&&= \frac{2b}{\sqrt{b^2+1}}
\end{align*}
- If \(a\) is fixed we have \(\left ( \frac{ab^2}{1+b^2}, -\frac{ba}{1+b^2} \right)\)
\begin{align*}
&& \frac{x}{y} &= - b \\
\Rightarrow && y &= \frac{a\frac{x}{y}}{1 + \frac{x^2}{y^2}} \\
\Rightarrow && y^2 \left ( 1 + \frac{x^2}{y^2} \right) &= ax \\
\Rightarrow && x^2-ax + y^2 &= 0 \\
\Rightarrow && \left (x - \frac{a}{2} \right)^2 + y^2 &= \frac{a^2}{4}
\end{align*}
Therefore we will end up with a circle centre \((\tfrac{a}{2}, 0)\) going through the origin.
- If \(ab = 1\), we have \(\left ( \frac{b}{1+b^2}, -\frac{1}{1+b^2} \right)\)
\begin{align*}
&& \frac{x}{y} &= -b \\
\Rightarrow && y &= -\frac{1}{1+\frac{x^2}{y^2}} \\
\Rightarrow && y + \frac{x^2}{y} &= - 1 \\
\Rightarrow && y^2 +y+ x^2 &= 0 \\
\Rightarrow && \left ( y + \frac12 \right)^2 + x^2 &= \frac14
\end{align*}
- If \(a\) is fixed we have \(\left ( \frac{ab^2}{1+b^2}, -\frac{ba}{1+b^2} \right)\)
\begin{align*}
&& \frac{x}{y} &= - b \\
\Rightarrow && y &= \frac{a\frac{x}{y}}{1 + \frac{x^2}{y^2}} \\
\Rightarrow && y^2 \left ( 1 + \frac{x^2}{y^2} \right) &= ax \\
\Rightarrow && x^2-ax + y^2 &= 0 \\
\Rightarrow && \left (x - \frac{a}{2} \right)^2 + y^2 &= \frac{a^2}{4}
\end{align*}
Therefore we will end up with a circle centre \((\tfrac{a}{2}, 0)\) going through the origin.
2010 Paper 1 Q7
Relative to a fixed origin \(O\), the points \(A\) and \(B\) have position vectors \(\bf{a}\) and \(\bf{b}\), respectively. (The points \(O\), \(A\) and \(B\) are not collinear.) The point \(C\) has position vector \(\bf c\) given by \[ {\bf c} =\alpha {\bf a}+ \beta {\bf b}\,, \] where \(\alpha\) and \(\beta\) are positive constants with \(\alpha+\beta<1\,\). The lines \(OA\) and \(BC\) meet at the point \(P\) with position vector \(\bf p\) and the lines \(OB\) and \(AC\) meet at the point \(Q\) with position vector \(\bf q\). Show that \[ {\bf p} =\frac{\alpha {\bf a} }{1-\beta}\,, \] and write down \(\bf q\) in terms of \(\alpha,\ \beta\) and \(\bf {b}\). Show further that the point \(R\) with position vector \(\bf r\) given by \[ {\bf r} =\frac{\alpha {\bf a} + \beta {\bf b}}{\alpha + \beta}\,, \] lies on the lines \(OC\) and \(AB\). The lines \(OB\) and \(PR\) intersect at the point \(S\). Prove that $ \dfrac{OQ}{BQ} = \dfrac{OS}{BS}\,$.
2009 Paper 3 Q1
The points \(S\), \(T\), \(U\) and \(V\) have coordinates \((s,ms)\), \((t,mt)\), \((u,nu)\) and \((v,nv)\), respectively. The lines \(SV\) and \(UT\) meet the line \(y=0\) at the points with coordinates \((p,0)\) and \((q,0)\), respectively. Show that \[ p = \frac{(m-n)sv}{ms-nv}\,, \] and write down a similar expression for \(q\). Given that \(S\) and \(T\) lie on the circle \(x^2 + (y-c)^2 = r^2\), find a quadratic equation satisfied by \(s\) and by \(t\), and hence determine \(st\) and \(s+t\) in terms of \(m\), \(c\) and \(r\). Given that \(S\), \(T\), \(U\) and \(V\) lie on the above circle, show that \(p+q=0\).
2003 Paper 2 Q5
The position vectors of the points \(A\,\), \(B\,\) and \(P\) with respect to an origin \(O\) are \(a{\bf i}\,\), \(b{\bf j}\,\) and \(l{\bf i}+m{\bf j}+n{\bf k}\,\), respectively, where \(a\), \(b\), and \(n\) are all non-zero. The points \(E\), \(F\), \(G\) and \(H\) are the midpoints of \(OA\), \(BP\), \(OB\) and \(AP\), respectively. Show that the lines \(EF\) and \(GH\) intersect. Let \(D\) be the point with position vector \(d{\bf k}\), where \(d\) is non-zero, and let \(S\) be the point of intersection of \(EF\) and \(GH.\) The point \(T\) is such that the mid-point of \(DT\) is \(S\). Find the position vector of \(T\) and hence find \(d\) in terms of \(n\) if \(T\) lies in the plane \(OAB\).
Solution: \(E = \langle \frac{a}{2}, 0,0 \rangle, F = \langle \frac{l}{2}, \frac{m+b}{2}, \frac{n}{2} \rangle, G = \langle 0, \frac{b}{2}, 0 \rangle, H = \langle \frac{a+l}{2}, \frac{m}{2}, \frac{n}{2} \rangle\) Note that the midpoint of \(EF\) and \(GH\) are both $\langle \frac{a+l}{4}, \frac{m+b}{4}, \frac{n}{4} \rangle$, so clearly they must intersect at this point. The vector we just found is \(S\), and \(\mathbf{t} = \mathbf{d} + 2(\mathbf{s}-\mathbf{d}) = 2\mathbf{s} - \mathbf{d}\). Therefore \(T = \langle \frac{a+l}{2}, \frac{m+b}{2}, \frac{n-2d}{2} \rangle\). If \(T\) lies in the plane \(OAB\) then \(n - 2d = 0\) ie \(d = \frac{n}{2}\)
1989 Paper 2 Q10
State carefully the conditions which the fixed vectors \(\mathbf{a,b,u}\) and \(\mathbf{v}\) must satisfy in order to ensure that the line \(\mathbf{r=a+}\lambda\mathbf{u}\) intersects the line \(\mathbf{r=b+\mu}\mathbf{v}\) in exactly one point. Find the two values of the fixed scalar \(b\) for which the planes with equations \[ \left.\begin{array}{c} x+y+bz=b+2\\ bx+by+z=2b+1 \end{array}\right\} \tag{*} \] do not intersect in a line. For other values of \(b\), express the line of intersection of the two planes in the form \(\mathbf{r=a}+\lambda\mathbf{u},\) where \(\mathbf{a\cdot u}=0\). Find the conditions which \(b\) and the fixed scalars \(c\) and \(d\) must satisfy to ensure that there is exactly one point on the line \[ \mathbf{r=}\left(\begin{array}{c} 0\\ 0\\ c \end{array}\right)+\mu\left(\begin{array}{c} 1\\ d\\ 0 \end{array}\right) \] whose coordinates satisfy both equations \((*)\).
Solution: There are two requirements (assuming they are lines not fixed points): 1. They cannot be parallel, ie \(\mathbf{u} \neq \lambda \mathbf{v}\) for any \(\lambda\) 2. They must lie in the same plane, ie \((\mathbf{b}-\mathbf{a})\cdot (\mathbf{u} \times \mathbf{v}) = 0\) The planes will not intersect in a line if they are either parallel and separate or parallel and the same. If \(b = 1\) or \(b=-1\) the planes are parallel. \begin{align*} && (x+y) + b z &= b+ 2\\ &&b(x+y) + z &= 2b + 1 \\ \Rightarrow && (1-b^2)z &= 2b+1 - b^2 -2b \\ &&&= 1-b^2 \\ \Rightarrow && z &= 1 \\ && x+ y &= 2 \\ \end{align*} Therefore our line is \(\mathbf{r} = \begin{pmatrix} 1+t \\ 1-t \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \) We must have: \(d \neq -1\) to ensure that the lines aren't parallel. We must also have: \begin{align*} 0 &= \left ( \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} -\begin{pmatrix} 0 \\ 0 \\ c \end{pmatrix}\right) \cdot \left ( \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 1 \\ d \\ 0 \end{pmatrix} \right) \\ &= \begin{pmatrix} 1 \\ 1 \\ 1-c \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ d+1 \end{pmatrix} \\ &= (1-c)(d+1) \end{align*} So \(c =1\)