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2024 Paper 3 Q4
- Show that if the acute angle between straight lines with gradients \(m_1\) and \(m_2\) is \(45^\circ\), then \[\frac{m_1 - m_2}{1 + m_1 m_2} = \pm 1.\]
- If \(p \neq q\), show that the tangents to the curve \(C\) at the points with \(x\)-coordinates \(p\) and \(q\) meet at a point with \(x\)-coordinate \(\frac{1}{2}(p+q)\). Find the \(y\)-coordinate of this point in terms of \(p\) and \(q\). Show further that any two tangents to the curve \(C\) which are at \(45^\circ\) to each other meet on the curve \((y+3a)^2 = 8a^2 + x^2\).
- Show that the acute angle between any two tangents to the curve \(C\) which meet on the curve \((y+7a)^2 = 48a^2 + 3x^2\) is constant. Find this acute angle.
2017 Paper 2 Q5
A curve \(C\) is determined by the parametric equations \[ x=at^2 \, , \; y = 2at\,, \] where \(a > 0\).
- Show that the normal to \(C\) at a point \(P\), with non-zero parameter \(p\), meets \(C\) again at a point \(N\), with parameter \(n\), where \[ n= - \left( p + \frac{2}{p} \right). \]
- Show that the distance \(\left| PN \right|\) is given by \[ \vert PN\vert^2 = 16a^2\frac{(p^2+1)^3}{p^4} \] and that this is minimised when \(p^2=2\,\).
- The point \(Q\), with parameter \(q\), is the point at which the circle with diameter \(PN\) cuts \(C\) again. By considering the gradients of \(QP\) and \(QN\), show that \[ 2 = p^2-q^2 + \frac{2q}p. \] Deduce that \(\left| PN \right|\) is at its minimum when \(Q\) is at the origin.
Solution:
- \(\,\) \begin{align*} && \frac{\d x}{\d t} &= 2at \\ && \frac{\d y}{\d t} &= 2a \\ \Rightarrow && \frac{\d y}{\d x} &= \frac1t \\ && -p &= \text{grad of normal} \\ &&&= \frac{y-2ap}{x-ap^2} \\ \Rightarrow && y &= -px + ap^3+2ap \\ && 2an &= -pan^2 + ap^3 + 2ap \\ \Rightarrow && 0 &= pan^2+2an-ap(2+p^2) \\ \Rightarrow && n &= p, -\left ( p + \frac2{p}\right) \\ \Rightarrow && n &= -\left ( p + \frac2{p}\right) \end{align*}
- \(\,\) \begin{align*} && |PN|^2 &= (ap^2-an^2)^2 +(2ap-2an)^2 \\ &&&= a^2(p-n)^2(p+n)^2+4a^2(p-n)^2 \\ &&&= a^2(p-n)^2((p+n)^2+4) \\ &&&= a^2\left(p+p+\frac2p \right)^2 \left ( \left ( -\frac2p\right)^2+4\right)\\ &&&= a^2\left(\frac{2p^2+2}p \right)^2 \left ( \frac{4}{p^2}+4\right)\\ &&&= 16a^2 \frac{(p^2+1)^3}{p^4} \\ \\ && \frac{\d |PN|^2}{\d p^2} &= 16a^2\frac{3(p^2+1)^2p^4-2(p^2+1)^3p^2}{p^8} \\ &&&= 16a^2(p^2+1)^2 \frac{3p^2-2(p^2+1)}{p^6} \\ &&&= 16a^2(p^2+1)^2 \frac{p^2-2}{p^6} \end{align*} Therefore minimized when \(p^2=2\) (clearly a minimum by considering behaviour as \(p^2 \to 0, \infty\))
- If \(PN\) is the diameter of \(PNQ\) then \(QP\) and \(QN\) are perpendicular, ie \begin{align*} && -1 &= \frac{2ap-2aq}{ap^2-aq^2} \cdot \frac{2aq-2an}{aq^2-an^2} \\ &&&= \frac{2}{p+q} \cdot \frac{2}{q+n} \\ &&&= \frac{2}{p+q} \cdot \frac{2}{q - p -\frac{2}{p}} \\ \Rightarrow && 4 &= (p+q)(p+\frac2{p}-q) \\ &&&= p^2-q^2 + \frac{2q}{p} + 2 \\ \Rightarrow && 2 &= p^2 - q^2 + \frac{2q}{p} \end{align*} Therefore \(q = 0 \Rightarrow p^2 = 2 \Rightarrow |PN|\) is at it's minimum.
2011 Paper 1 Q1
- Show that the gradient of the curve \(\; \dfrac a x + \dfrac by =1\), where \(b\ne0\), is \(\; -\dfrac{ay^2}{bx^2}\,\). The point \((p,q)\) lies on both the straight line \(ax+by=1\) and the curve \(\dfrac a x + \dfrac by =1\,\), where \(ab\ne0\). Given that, at this point, the line and the curve have the same gradient, show that \( p=\pm q\,\). Show further that either \((a-b)^2 =1\,\) or \((a+b)^2 =1\,\).
- Show that if the straight line \(ax+by=1\), where \(ab\ne0\), is a normal to the curve \(\dfrac a x - \dfrac by =1\), then \(a^2-b^2 = \frac12\,\).
Solution:
- \(\,\) \begin{align*} && 1 &= \frac{a}{x} + \frac{b}{y} \\ \frac{\d}{\d x}: && 0 &= -\frac{a}{x^2} - \frac{b}{y^2} \frac{\d y}{\d x} \\ \Rightarrow && \frac{\d y}{\d x} &= -\frac{ay^2}{bx^2} \\ \\ (p,q): && -\frac{aq^2}{bp^2} &= -\frac{a}{b} \\ \Rightarrow && p^2 &= q^2 \\ \Rightarrow && p &= \pm q \\ \\ \Rightarrow && ap \pm b p &= 1 \\ \Rightarrow && (a\pm b)p &= 1 \\ \Rightarrow && \frac{a}{p} \pm \frac{b}{p} &= 1 \\ \Rightarrow && (a \pm b)\frac{1}{p} &= 1 \\ \Rightarrow && (a \pm b)^2 &= 1 \end{align*}
- \(\,\) \begin{align*} && 1 &= \frac{a}{x} - \frac{b}{y} \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{ay^2}{bx^2} \\ \Rightarrow && \frac{aq^2}{bp^2} &= \frac{b}{a} \\ \Rightarrow && aq &= \pm bp \\ \Rightarrow && 1 &= \frac{a}{p} - \frac{b}{q} \\ &&&= \frac{aq-bp}{pq} \\ \Rightarrow && aq &= -bp \\ \Rightarrow && 1 &= \frac{2aq}{pq} \\ \Rightarrow && p &= 2a \\ \Rightarrow && q &= -2b \\ \Rightarrow && 1 &= 2a^2-2b^2 \\ \Rightarrow && \frac12 &= a^2-b^2 \end{align*}
2010 Paper 1 Q3
Show that \[ \sin(x+y) -\sin(x-y) = 2 \cos x \, \sin y \] and deduce that \[ \sin A - \sin B = 2 \cos \tfrac12 (A+B) \, \sin\tfrac12 (A-B) \,. \] Show also that \[ \cos A - \cos B = -2 \sin \tfrac12(A+B) \, \sin\tfrac12(A-B)\,. \] The points \(P\), \(Q\), \(R\) and \(S\) have coordinates \(\left(a\cos p,b\sin p\right)\), \(\left(a\cos q,b\sin q\right)\), \(\left(a\cos r,b\sin r\right)\) and \(\left(a\cos s,b\sin s\right)\) respectively, where \(0\le p < q < r < s <2\pi\), and \(a\) and \(b\) are positive. Given that neither of the lines \(PQ\) and \(SR\) is vertical, show that these lines are parallel if and only if \[ r+s-p-q = 2\pi\,. \]
Solution: \begin{align*} && \sin(x+y) - \sin(x-y) &= \sin x \cos y + \cos x \sin y - (\sin x \cos y - \cos x \sin y )\\ &&&= 2 \cos x \sin y \\ \\ && A &= x+y \\ && B &= x - y \\ \Rightarrow && x = \frac12(A+B) &\quad y = \frac12(A-B) \\ \Rightarrow && \sin A - \sin B &= 2 \cos \tfrac12(A+B) \sin \tfrac12(A-B) \\ \\ && \cos (x+y) - \cos (x-y) &= \cos x \cos y - \sin x \sin y -(\cos x \cos y + \sin x \sin y ) \\ &&&= -2 \sin x \sin y \\ \Rightarrow && \cos A - \cos B &= - 2 \sin \tfrac12 (A+B) \sin \tfrac12 (A-B) \end{align*} \begin{align*} && \text{Gradient of }PQ &= \frac{b \sin q - b \sin p}{a \cos q - a \cos p } \\ && \text{Gradient of }SR &= \frac{b \sin s - b \sin r}{a \cos s - a \cos r} \\ PQ \parallel SR \Rightarrow && \frac{b \sin q - b \sin p}{a \cos q - a \cos p } &= \frac{b \sin s - b \sin r}{a \cos s - a \cos r} \\ \Rightarrow && (\sin q - \sin p)(\cos s - \cos r) &= (\sin s - \sin r)(\cos q - \cos r) \\ \Rightarrow && -4 \cos \tfrac12(p+q) \sin\tfrac12(q-p) \sin \tfrac12(s+r) \sin \tfrac12(s-r) &= -4 \cos \tfrac12(s+r) \sin \tfrac12(s-r) \sin \tfrac12 (p+q) \sin\tfrac12 (q-p) \\ \Rightarrow && 0 &= \sin \tfrac12(s-r)\sin\tfrac12(p-q) \left ( \cos \tfrac12(p+q)\sin \tfrac12(s+r) - \sin \tfrac12 (p+q)\cos \tfrac12(s+r) \right) \\ &&&= \sin \tfrac12(s-r)\sin\tfrac12(p-q) \sin \left ( \frac12 (s+r -(p+q))\right) \end{align*} Since \(s \neq r\) and \(p \neq q\) (neither line vertical) we must have \(\frac12 (s+r -(p+q)) = n \pi \Rightarrow s+r - p - q = 0, 2\pi, 4\pi, \cdots\) but given the range constraints it must be \(2 \pi\)
2006 Paper 3 Q6
Show that in polar coordinates the gradient of any curve at the point \((r,\theta)\) is \[ \frac{ \ \ \dfrac{\d r }{\d\theta} \tan\theta + r \ \ } { \dfrac{\d r }{\d\theta} -r\tan\theta}\,. \] \noindent
1998 Paper 2 Q6
Two curves are given parametrically by \[ x_{1}=(\theta+\sin\theta),\qquad y_{1}=(1+\cos\theta),\tag{1} \]and \[ x_{2}=(\theta-\sin\theta),\qquad y_{1}=-(1+\cos\theta),\tag{2} \] Find the gradients of the tangents to the curves at the points where \(\theta= \pi/2\) and \(\theta=3\pi/2\). Sketch, using the same axes, the curves for \(0\le\theta \le 2\pi\). Find the equation of the normal to the curve (1) at the point with parameter \(\theta\). Show that this normal is a tangent to the curve (2).
1990 Paper 1 Q3
Given a curve described by \(y=\mathrm{f}(x)\), and such that \(y\geqslant0\), a push-off of the curve is a new curve obtained as follows: for each point \((x,\mathrm{f}(x))\) with position vector \(\mathbf{r}\) on the original curve, there is a point with position vector \(\mathbf{s}\) on the new curve such that \(\mathbf{s-r}=\mathrm{p}(x)\mathbf{n},\) where \(\mathrm{p}\) is a given function and \(\mathbf{n}\) is the downward-pointing unit normal to the original curve at \(\mathbf{r}\).
- For the curve \(y=x^{k},\) where \(x>0\) and \(k\) is a positive integer, obtain the function \(\mathrm{p}\) for which the push-off is the positive \(x\)-axis, and find the value of \(k\) such that, for all points on the original curve, \(\left|\mathbf{r}\right|=\left|\mathbf{r-s}\right|\).
- Suppose that the original curve is \(y=x^{2}\) and \(\mathrm{p}\) is such that the gradient of the curves at the points with position vectors \(\mathbf{r}\) and \(\mathbf{s}\) are equal (for every point on the original curve). By writing \(\mathrm{p}(x)=\mathrm{q}(x)\sqrt{1+4x^{2}},\) where \(\mathrm{q}\) is to be determined, or otherwise, find the form of \(\mathrm{p}\).
Solution:
- Suppose we have \(y = x^k\), then the tangent at \((t,t^k)\) has gradient \(\frac{\d y}{\d x} = kx^{k-1} = kt^{k-1}\) and the normal has gradient \(-\frac1k x^{1-k}\). For the push-off to be the positive \(x\)-axis, we need \(p(x)\) to be the length of the line. The line will have the equation: \begin{align*} && \frac{y - t^k}{x - t} &= -\frac1k t^{1-k} \\ y = 0: && x-t &= \frac{kt^k}{t^{1-k}} \\ && x& =t + kt^{2k-1} \end{align*} The distance from \((t,t^k)\) to \((t+kt^{2k-1},0)\) is \(\sqrt{t^{2k} + k^2t^{4k-2}} = t^k \sqrt{1+k^2t^{2k-2}} = y \sqrt{1+k^2 \frac{y^2}{x^2}} = \frac{y}{x} \sqrt{x^2 + k^2y^2}\), ie \(p(x) = \frac{y}{x}\sqrt{x^2+k^2y^2}\) If \(\left|\mathbf{r}\right|=\left|\mathbf{r-s}\right|\), then we need \(\sqrt{x^2+y^2} = \frac{y}{x}\sqrt{x^2+ky^2}\), but clearly this is satisfied when \(k = 1\).
- The points are \((t, t^2)\) and the normal has graident \(-\frac{1}{2t}\), the normal vector is \(\frac{1}{\sqrt{1+\frac{1}{4t^2}}}\binom{1}{-\frac1{2t}} = \frac{1}{\sqrt{4t^2+1}} \binom{2t}{-1}\). If we write \(p(x) = q(x)\sqrt{4x^2+1}\) then the the new points are at \(\binom{t+q(t)2t}{t^2-\q(t)}\) an the gradient will be \(\frac{2t-q'(t)}{1+q'(t)2t+2q(t)}\). We need it to be the case that \begin{align*} && 2t &= \frac{2t-q'(t)}{1+2q(t)+2tq'(t)} \\ \Rightarrow && 4tq(t) &= -q'(t)(1+4t^2) \\ \Rightarrow && \frac{q'(t)}{q(t)} &= -\frac{4t}{1+4t^2} \\ \Rightarrow && \ln q(t) &= -\frac12 \ln(1+4t^2)+C \\ \Rightarrow && q(t) &= A(1+4t^2)^{-1/2} \\ \Rightarrow && p(x) &= A \end{align*} So the push-off's are constants.
1989 Paper 2 Q5
- Show that in polar coordinates, the gradient of any curve at the point
\((r,\theta)\) is
\[
\left.\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}\tan\theta+r\right)\right/\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}-r\tan\theta\right).
\]
- A mirror is designed so that any ray of light which hits one side of the mirror and which is parallel to a certain fixed line \(L\) is reflected through a fixed point \(O\) on \(L\). For any ray hitting the mirror, the normal to the mirror at the point of reflection bisects the angle between the incident ray and the reflected ray, as shown in the figure. Prove that the mirror intersects any plane containing \(L\) in a parabola.
Solution:
- Suppose our curve is \(r(\theta)\), then \(y = r \sin \theta, x = r \cos \theta\) and \begin{align*} && \frac{\d y}{\d \theta} &= \frac{\d r}{\d \theta} \sin \theta + r \cos \theta \\ && \frac{\d x}{\d \theta} &= \frac{\d r}{\d \theta} \cos \theta - r \sin \theta \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{\d y}{\d \theta} \Bigg / \frac{\d x}{\d \theta} \\ &&&= \frac{\frac{\d r}{\d \theta} \sin \theta + r \cos \theta}{\frac{\d r}{\d \theta} \cos \theta - r \sin \theta} \\ &&&= \frac{\frac{\d r}{\d \theta} \tan\theta + r }{\frac{\d r}{\d \theta} - r \tan\theta} \end{align*} as required.
- Set up a system of polar coordinates such that the origin is at \(O\) and all points in the plane containing \(L\) are represented by \((r, \theta)\). The constraint we have is that the angle of the normal, is \(\frac12 \theta\). Let \(\tan \tfrac12 \theta = t\), then \(\tan \theta = \frac{2t}{1-t^2}\) \begin{align*} && \tan \frac12 \theta &= -\frac{\frac{\d r}{\d \theta} - r \tan\theta}{\frac{\d r}{\d \theta} \tan\theta + r } \\ \Rightarrow && t &= -\frac{r'-r\frac{2t}{1-t^2}}{r' \frac{2t}{1-t^2}+r} \\ &&&= \frac{2tr-(1-t^2)r'}{2tr'+(1-t^2)r} \\ \Rightarrow && (2t^2+1-t^2)r' &= (2t-t+t^3)r \\ && (1+t^2)r' &= t(t^2+1) r \\ \Rightarrow && r' &= t r \\ \Rightarrow && \frac{\d r}{\d \theta} &= \tan \tfrac12 \theta r \\ \Rightarrow && \int \frac1r \d r &= \int \tan \frac12 \theta \d \theta \\ && \ln r &= -2\ln \cos \tfrac12 \theta+C \\ \Rightarrow && r\cos^2 \frac12 \theta &= C \\ \Rightarrow && r + r\cos \theta &= D \\ \Rightarrow && r &= D-x \\ \Rightarrow && x^2 + y^2 &= D^2 - 2Dx + x^2 \\ \Rightarrow && y^2 &= D^2-2Dx \end{align*} Therefore it is a parabola