Problems

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Solution: \begin{align*} \sum_{r=0}^{n-1} \e^{2i(\alpha + r\pi/n)} &= e^{2i\alpha} \sum_{r=0}^{n-1} \left (\e^{2i\pi/n} \right)^r \\ &= e^{2i\alpha} \frac{1-\left (\e^{2i\pi/n} \right)^n}{1-\e^{2i\pi/n} } \\ &= 0 \end{align*} \(d = s + r \cos \theta\) ie \(s = d - r \cos \theta\) Therefore \(d = \frac{r}{k} + r \cos \theta \Rightarrow r = \frac{kd}{1+k \cos \theta}\). The \(l_j\) will come from \(r(\alpha + \frac{j \pi}{n} )+r(\alpha + \pi + \frac{j \pi}{n} )\) \begin{align*} && l_j &= r(\alpha + \frac{(j-1) \pi}{n} )+r(\alpha + \pi + \frac{(j-1) \pi}{n} ) \\ &&&= \frac{kd}{1+k \cos \left ( \alpha + \frac{(j-1) \pi}{n}\right)}+\frac{kd}{1+k \cos \left ( \alpha+\pi+ \frac{(j-1) \pi}{n}\right)}\\ &&&= \frac{kd}{1+k \cos \left ( \alpha + \frac{(j-1) \pi}{n}\right)}+\frac{kd}{1-k \cos \left ( \alpha+ \frac{(j-1) \pi}{n}\right)}\\ &&&= \frac{2kd}{1-k^2 \cos^2 \left ( \alpha + \frac{(j-1) \pi}{n}\right)}\\ \Rightarrow && \sum_{j=1}^n \frac 1 {l_j} &= \sum_{j=0}^{n-1} \frac{1-k^2 \cos^2 \left ( \alpha + \frac{j \pi}{n}\right)}{2kd} \\ &&&= \frac{n}{2kd}-\frac{k^2}{2kd} \sum_{j=0}^{n-1} \cos^2 \left ( \alpha + \frac{j \pi}{n}\right) \\ &&&= \frac{n}{2kd}-\frac{k^2}{2kd} \sum_{j=0}^{n-1} \frac{1+ \cos \left ( 2\alpha + \frac{2j \pi}{n}\right)}{2} \\ &&&= \frac{n}{2kd}-\frac{nk^2}{2kd}-\frac{k^2}{4kd} \sum_{j=0}^{n-1}\cos \left ( 2\alpha + \frac{2j \pi}{n}\right) \\ &&&= \frac{n}{2kd}-\frac{nk^2}{2kd}-\frac{k^2}{4kd} \underbrace{\textrm{Re} \left ( \sum_{j=0}^{n-1}e^{ 2i(\alpha + \frac{j \pi}{n})} \right)}_{=0} \\ &&&= \frac{n}{2kd} - \frac{nk^2}{4kd} \\ &&&= \frac{n(2-k^2)}{4kd} \end{align*}

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Solution: Find the gradient of the tangent of the ellipse at \(P\): \begin{align*} && \frac{2x}{a^2} + \frac{2y}{b^2} \frac{\d y}{\d x} &= 0 \\ \Rightarrow && \frac{\d y}{\d x} &= - \frac{2xb^2}{2ya^2} \\ &&&=- \frac{a \cos \theta b^2}{b \sin \theta a^2} \\ &&&=-\frac{b}{a} \cot \theta \end{align*} Therefore the gradient of \(ON\) is \(\frac{a}{b} \tan \theta\). \begin{align*} && y &= \frac{a}{b} \tan \theta x \\ && \frac{y-0}{x-(-ea)} &= \frac{b\sin \theta-0}{a\cos \theta -(-ea)} \\ && y &= \frac{b \sin \theta}{a(e+\cos \theta)}(x+ea) \\ \Rightarrow && y &= \frac{b \sin \theta}{a(\cos \theta+e)}\frac{b}{a} \cot \theta y+ \frac{eb \sin \theta}{\cos \theta + e} \\ &&&= \frac{b^2 \cos \theta}{a^2(\cos \theta +e)}y + \frac{eb \sin \theta}{\cos \theta + e} \\ \Rightarrow && (\cos \theta+e)y &= (1-e^2)\cos \theta y +eb \sin \theta\\ && e(1+e\cos \theta)y &= eb \sin \theta \\ \Rightarrow && y &= \frac{b \sin \theta}{1+e\cos \theta} \\ && x &= \frac{b \sin \theta}{1+e\cos \theta} \frac{b}{a} \cot \theta \\ &&&= \frac{b^2 \cos \theta}{a(1+e\cos \theta)} \end{align*} Therefore \(\displaystyle T\left (\frac{b^2 \cos \theta}{a(1+e\cos \theta)}, \frac{b \sin \theta}{1+e\cos \theta} \right)\). Finally, we can look at the distance of \(T\) from \(S\) \begin{align*} && d^2 &= \left (\frac{b^2 \cos \theta}{a(1+e\cos \theta)}-(-ea) \right)^2 + \left (\frac{b \sin \theta}{1+e\cos \theta} -0\right)^2 \\ &&&= \frac{\left (b^2 \cos \theta+ea^2(1+e\cos\theta)\right)^2 + \left ( ab \sin \theta\right)^2}{a^2(1+e\cos \theta)^2} \\ &&&= \frac{b^4\cos^2\theta+e^2a^4(1+e\cos\theta)^2+2ea^2b^2(1+e\cos\theta)+a^2b^2\sin^2\theta}{a^2(1+e\cos\theta)^2} \\ &&&= \frac{a^4(1-e^2)^2\cos^2\theta+e^2a^4(1+e\cos\theta)^2+2ea^2a^2(1-e^2)(1+e\cos\theta)+a^4(1-e^2)\sin^2\theta}{a^2(1+e\cos\theta)^2} \\ &&&= a^2 \left ( \frac{(1-e^2)^2\cos^2\theta+e^2(1+e\cos\theta)^2+2e(1-e^2)(1+e\cos\theta)+(1-e^2)(1-\cos^2\theta)}{(1+e\cos\theta)^2} \right) \\ &&&= a^2 \left ( \frac{e^2(1+e\cos\theta)^2+(1-e^2)((1-e^2)\cos^2\theta+2e(1+e\cos\theta)+(1-\cos^2\theta))}{(1+e\cos\theta)^2} \right) \\ &&&= a^2 \left ( \frac{e^2(1+e\cos\theta)^2+(1-e^2)(1+e\cos\theta)^2}{(1+e\cos\theta)^2} \right) \\ &&&= a^2 \end{align*} Therefore a circle radius \(a\) centre \(S\).

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Solution:

\begin{align*} && \ell &= r(1 + e \cos \theta) \\ \Rightarrow && 0 &= \frac{\d r}{\d \theta}(1 + e \cos \theta) - re \sin \theta \\ \Rightarrow && \frac{\d r}{\d \theta} &= \frac{re \sin \theta}{1+e \cos \theta} \end{align*} Suppose we consider the \((x',y')\) plane, which is essentially the \(x-y\) plan rotated by \(45^\circ\), then we would have \begin{align*} && \frac{\d y'}{ \d x'} &= \frac{\frac{\d y'}{\d \theta}}{\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{re \sin \theta}{1+e \cos \theta} \sin \theta + r\cos \theta}{\frac{re \sin \theta}{1+e \cos \theta} \cos\theta -r\sin \theta} \\ &&&= \frac{re\sin^2 \theta+r \cos \theta(1+e \cos \theta)}{re\sin \theta \cos \theta -r \sin \theta (1+e \cos \theta)} \\ &&&= \frac{\cos \theta + e \cos^2 \theta+e \sin^2 \theta}{-\sin \theta} \\ &&&= \frac{\cos \theta + e}{-\sin \theta} \end{align*} Since our frame is rotated by \(45^\circ\) we need to consider the appropriate gradient for this. We know that \(m = \tan \theta\) so \(m' = \tan (\theta+45^{\circ}) = \frac{1+m}{1-m}\) therefore we should have \begin{align*} && \frac{\d y}{ \d x} &= \frac{1+\frac{\cos \theta + e}{-\sin \theta}}{1-\frac{\cos \theta + e}{-\sin \theta}} \\ &&&= \frac{\cos \theta - \sin \theta + e}{-\sin \theta - \cos \theta-e} \\ &&&= \frac{\sin \theta - \cos \theta -e}{\sin \theta + \cos \theta +e} \end{align*} As required. The tangents at those points are parallel, therefore \begin{align*} && \frac{\cos \alpha+e}{\sin \alpha} &= \frac{\cos \beta+e}{\sin \beta} \\ \Rightarrow && \frac{\frac{1-\tan^2 \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}}+e}{\frac{2\tan \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}}} &= \frac{\frac{1-\tan^2 \frac{\beta}{2}}{1+\tan^2 \frac{\beta}{2}}+e}{\frac{2\tan \frac{\beta}{2}}{1+\tan^2 \frac{\beta}{2}}} \\ && \frac{1-\tan^2 \frac{\alpha}{2}+e(1+\tan^2\frac{\alpha}{2})}{2\tan\frac{\alpha}{2}} &= \frac{1-\tan^2 \frac{\beta}{2}+e(1+\tan^2\frac{\beta}{2})}{2\tan\frac{\beta}{2}} \\ && \frac{(1+e)+(e-1)\tan^2 \frac{\alpha}{2}}{2\tan \frac{\alpha}{2}} &= \frac{(1+e)+(e-1)\tan^2 \frac{\beta}{2}}{2\tan \frac{\beta}{2}} \\ && \frac{(1+e)}{\tan\frac{\alpha}2} - (1-e)\tan\frac{\alpha}2 &= \frac{(1+e)}{\tan\frac{\beta}2} - (1-e)\tan\frac{\beta}2 \end{align*} ie both \(\tan \frac{\alpha}{2}\) and \(\tan \frac{\beta}{2}\) are roots of a quadratic of the form \((1-e)x^2-cx-(1+e)\) but this means the product of the roots is \(-\frac{1+e}{1-e}\)

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Solution: \begin{align*} && \frac1{r} &= A \cos \theta + B \cos (\theta - \gamma) \\ &&&= A \cos \theta + B \cos \theta \cos \gamma + B \sin \theta \sin \gamma \\ &&&= (A+B \cos \gamma) \cos \theta + B \sin \gamma \sin \theta \\ \Longleftrightarrow && 1 &= (A+B \cos \gamma) x + B \sin \gamma y \end{align*} So if we choose \(B = \frac{m}{\sin \gamma}\) and \(A = l-m \cot \gamma\) we have the desired result. \begin{align*} && \frac{1 + e \cos (\gamma -\delta)}a &= A \cos (\gamma - \delta) + B \cos (\gamma - \delta - \gamma) \\ &&&= A \cos(\gamma-\delta) +B \cos \delta\\ && \frac{1 + e \cos (\gamma +\delta)}{a} &= A \cos (\gamma + \delta) + B \cos (\gamma + \delta - \gamma) \\ &&&= A \cos(\gamma + \delta) + B \cos \delta\\ \Rightarrow && \frac1{r} &= \frac{e}{a} \cos \theta + \frac{1}{a \cos \delta} \cos (\theta - \gamma) \\ \lim{\delta \to 0} &&\frac1{r} &= \frac{e}{a} \cos \theta+ \frac{1}{a} \cos (\theta - \gamma) \end{align*} Suppose we have have points \(L\) and \(M\) with \(\theta = \gamma_L, \gamma_M\) then our tangents are: \begin{align*} && \frac{a}{r} &= \cos \theta + \cos (\theta - \gamma_L) \\ && \frac{a}{r} &= \cos \theta + \cos (\theta - \gamma_M) \\ \Rightarrow && 0 &= \cos (\theta - \gamma_L) -\cos(\theta - \gamma_M) \\ &&&= - 2 \sin \frac{(\theta - \gamma_L)+(\theta - \gamma_M)}{2} \sin \frac{(\theta - \gamma_L)-(\theta - \gamma_M)}{2} \\ &&&= -2 \sin \left ( \theta - \frac{\gamma_L+\gamma_M}2 \right) \sin \left ( \frac{\gamma_M - \gamma_L}{2}\right) \\ \Rightarrow && \theta &= \frac{\gamma_L+\gamma_M}{2} \end{align*} Therefore clearly \(ST\) bisects \(LSM\). The line \(LM\) can be seen as the chord from the points \(\frac{\gamma_L+\gamma_M}{2} \pm \frac{\gamma_L-\gamma_M}{2}\), so the line is: \begin{align*} && \frac{a}{r} &= e \cos \theta + \frac{1}{\cos \left ( \frac{\gamma_L-\gamma_M}{2}\right)} \cos \left (\theta - \frac{\gamma_L+\gamma_M}{2} \right) \end{align*} and we want the point on the line where \(\theta =\frac{\gamma_L+\gamma_M}{2}\) so \begin{align*} && \frac{a}{r} &= e \cos \left ( \frac{\gamma_L+\gamma_M}{2} \right) + \frac{1}{\cos \left ( \frac{\gamma_L-\gamma_M}{2}\right)} \\ \Rightarrow && r &= \frac{a}{e \cos \left ( \frac{\gamma_L+\gamma_M}{2} \right) + \frac{1}{\cos \left ( \frac{\gamma_L-\gamma_M}{2}\right)}} \end{align*}