1 problem found
Show that, for a given constant \(\gamma\) \((\sin\gamma\neq0)\) and with suitable choice of the constants \(A\) and \(B\), the line with cartesian equation \(lx+my=1\) has polar equations \[ \frac{1}{r}=A\cos\theta+B\cos(\theta-\gamma). \] The distinct points \(P\) and \(Q\) on the conic with polar equations \[ \frac{a}{r}=1+e\cos\theta \] correspond to \(\theta=\gamma-\delta\) and \(\theta=\gamma+\delta\) respectively, and \(\cos\delta\neq0.\) Obtain the polar equation of the chord \(PQ.\) Hence, or otherwise, obtain the equation of the tangent at the point where \(\theta=\gamma.\) The tangents at \(L\) and \(M\) to a conic with focus \(S\) meet at \(T.\) Show that \(ST\) bisects the angle \(LSM\) and find the position of the intersection of \(ST\) and \(LM\) in terms of your chosen parameters for \(L\) and \(M.\)
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*}