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2023 Paper 3 Q1
The distinct points \(P(2ap,\, ap^2)\) and \(Q(2aq,\, aq^2)\) lie on the curve \(x^2 = 4ay\), where \(a > 0\).
- Given that \[(p+q)^2 = p^2q^2 + 6pq + 5\,,\tag{\(*\)}\] show that the line through \(P\) and \(Q\) is a tangent to the circle with centre \((0,\, 3a)\) and radius \(2a\).
- Show that, for any given value of \(p\) with \(p^2 \neq 1\), there are two distinct real values of \(q\) that satisfy equation \((*)\). Let these values be \(q_1\) and \(q_2\). Find expressions, in terms of \(p\), for \(q_1 + q_2\) and \(q_1 q_2\).
- Show that, for any given value of \(p\) with \(p^2 \neq 1\), there is a triangle with one vertex at \(P\) such that all three vertices lie on the curve \(x^2 = 4ay\) and all three sides are tangents to the circle with centre \((0,\, 3a)\) and radius \(2a\).
2013 Paper 1 Q5
The point \(P\) has coordinates \((x,y)\) which satisfy \[ x^2+y^2 + kxy +3x +y =0\,. \]
- Sketch the locus of \(P\) in the case \(k=0\), giving the points of intersection with the coordinate axes.
- By factorising \(3x^2 +3y^2 +10xy\), or otherwise, sketch the locus of \(P\) in the case \(k=\frac{10}{3}\,\), giving the points of intersection with the coordinate axes.
- In the case \(k=2\), let \(Q\) be the point obtained by rotating \(P\) clockwise about the origin by an angle~\(\theta\), so that the coordinates \((X,Y)\) of \(Q\) are given by \[ X=x\cos\theta +y\sin\theta\,, \ \ \ \ Y= -x\sin\theta + y\cos\theta\,. \] Show that, for \(\theta =45^\circ\), the locus of \(Q\) is \( \sqrt2 Y= (\sqrt2 X+1 )^2 - 1 .\) Hence, or otherwise, sketch the locus of \(P\) in the case \(k=2\), giving the equation of the line of symmetry.
Solution:
- \(k = 0\), we have \(x^2 + y^2 + 3x + y = 0\), ie \((x+\tfrac32)^2+(y+\tfrac12)^2 = \frac{10}{4}\).
- \(3x^2 + 3y^2 +10xy = (3x+y)(x+3y)\) so \(x^2 + y^2 + \tfrac{10}3xy + 3x+y = (3x+y)(\frac{x+3y}{3}+1) = 0\) so we have the line pair \(3x +y =0\), \(x+3y + 3 = 0\)
- If \(k = 2\) then \((x+y)^2 + (x+y)+2x = 0\). If \(\theta = 45^\circ\) then \( X = \frac1{\sqrt{2}}(x+y), Y = \frac{1}{\sqrt{2}}(y-x)\), ie \(x+y = \sqrt{2}X\) and \(x = \frac{1}{\sqrt2}(X-Y)\), so our equation is:
\begin{align*}
0 &= 2X^2 + \sqrt{2}X + \sqrt{2}(X-Y) \\
&= (\sqrt{2}X + 1)^2 - 1 - \sqrt{2} Y
\end{align*}
which would be a parabola with line of symmetry \(X = -\frac{1}{\sqrt{2}}\).
However, we are actually looking at that parabola rotated by \(45^\circ\) anticlockwise.
1993 Paper 2 Q10
Verify that if \[ \mathbf{P}=\begin{pmatrix}1 & 2\\ 2 & -1 \end{pmatrix}\qquad\mbox{ and }\qquad\mathbf{A}=\begin{pmatrix}-1 & 8\\ 8 & 11 \end{pmatrix} \] then \(\mathbf{PAP}\) is a diagonal matrix. Put $\mathbf{x}=\begin{pmatrix}x\\ y \end{pmatrix}\( and \)\mathbf{x}_{1}=\begin{pmatrix}x_{1}\\ y_{1} \end{pmatrix}.$ By writing \[ \mathbf{x}=\mathbf{P}\mathbf{x}_{1}+\mathbf{a} \] for a suitable vector \(\mathbf{a},\) show that the equation \[ \mathbf{x}^{\mathrm{T}}\mathbf{Ax}+\mathbf{b}^{\mathrm{T}}\mathbf{x}-11=0, \] where $\mathbf{b}=\begin{pmatrix}18\\ 6 \end{pmatrix}\( and \) \mathbf{x}^{\mathrm{T}} \( is the transpose of \)\mathbf{x},$ becomes \[ 3x_{1}^{2}-y_{1}^{2}=c \] for some constant \(c\) (which you should find).
Solution: \begin{align*} \mathbf{PAP} &= \begin{pmatrix}1 & 2\\ 2 & -1 \end{pmatrix}\begin{pmatrix}-1 & 8\\ 8 & 11 \end{pmatrix}\begin{pmatrix}1 & 2\\ 2 & -1 \end{pmatrix} \\ &= \begin{pmatrix}1 & 2\\ 2 & -1 \end{pmatrix}\begin{pmatrix}15 & -10\\ 30 & 5 \end{pmatrix} \\ &= \begin{pmatrix}75 & 0\\ 0 & -25 \end{pmatrix} \end{align*} Which is diagonal as required. Letting \(\mathbf{x}=\mathbf{P}\mathbf{x}_{1}+\mathbf{a}\) \begin{align*} && \mathbf{x}^{\mathrm{T}}\mathbf{Ax}+\mathbf{b}^{\mathrm{T}}\mathbf{x}-11&=0 \\ \Leftrightarrow && (\mathbf{P}\mathbf{x}_{1}+\mathbf{a})^T\mathbf{A}(\mathbf{P}\mathbf{x}_{1}+\mathbf{a}) + \mathbf{b}^T(\mathbf{P}\mathbf{x}_{1}+\mathbf{a}) - 11 &= 0 \\ \Leftrightarrow && \mathbf{x}_{1}^T\mathbf{PAP}\mathbf{x}_1 + \mathbf{x}_{1}^T\mathbf{PAa} + \mathbf{a}^T\mathbf{AP}\mathbf{x}_{1}+\mathbf{a}^T\mathbf{Aa} + \mathbf{b}^T(\mathbf{P}\mathbf{x}_{1}+\mathbf{a}) - 11 &= 0 \\ \Leftrightarrow && \mathbf{x}_{1}^T\mathbf{PAP}\mathbf{x}_1 +(2\mathbf{a}^T\mathbf{A}+\mathbf{b}^T)\mathbf{P}\mathbf{x}_{1}+\mathbf{a}^T\mathbf{Aa} + \mathbf{b}^T\mathbf{a} - 11 &= 0 \\ \end{align*} It would be nice if we picked \(\mathbf{a}\) such that \(2\mathbf{a}^T\mathbf{A}+\mathbf{b}^T = 0\), if \(\mathbf{a} = \begin{pmatrix} a_1 \\a_2 \end{pmatrix}\) then this equation becomes: \begin{align*} && 2\begin{pmatrix}-a_1 + 8a_2 & 8a_1+11a_2 \end{pmatrix} + \begin{pmatrix}18 & 6 \end{pmatrix} &= 0 \\ \Rightarrow && a_1 = 1, a_2 = -1 \end{align*} So our equation is now \begin{align*} && \mathbf{x}_{1}^T\mathbf{PAP}\mathbf{x}_1 +(2\mathbf{a}^T\mathbf{A}+\mathbf{b}^T)\mathbf{P}\mathbf{x}_{1}+\mathbf{a}^T\mathbf{Aa} + \mathbf{b}^T\mathbf{a} - 11 &= 0 \\ \Leftrightarrow && \mathbf{x}_{1}^T\mathbf{PAP}\mathbf{x}_1-6 +12 - 11 &= 0 \\ \Leftrightarrow && 25(3x_1^2 - y_1^2) &= 5 \\ \Leftrightarrow && 3x_1^2 - y_1^2 &= \frac{1}{5} \end{align*}