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Hyperbola WO.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=1cm]{geometry} \begin{document} \noindent {\Large Hyperbola \--- The Working Out} \begin{align*} \text{In the following,}& \quad a>0, \quad b>0,\\ &\quad\text{$P(x_1,y_1)$ or $P(a\sec\theta_1,b\tan\theta_1)$ is on the hyperbola, so is}\\ &\quad\text{$Q(x_2,y_2)$ or $Q(a\sec\theta_2,b\tan\theta_2)$, and}\\ &\quad\text{$T(x_0,y_0)$ or $T(a\sec\theta_0,b\tan\theta_0)$ is the intersection of the}\\ &\qquad\qquad\text{two tangents from $P$ and $Q$.} \end{align*} % % Hyperbola (Cartesian) % \begin{align*} \text{{\bf Hyperbola} (Cartesian)}\qquad&\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\:,\quad b^2 x^2-a^2 y^2=a^2 b^2\:,\quad y=\pm b\sqrt{\frac{x^2}{a^2}-1} \end{align*} \begin{align*} \text{Derivative:}&\quad\bf\frac{b^2}{a^2}\frac{x}{y}\\ &\quad b^2 x^2-a^2 y^2=a^2 b^2\\ &\quad 2b^2 x\:dx-2a^2 y\:dy=0\\ &\quad\not{2}a^2 y\:dy=\not{2}b^2 x\:dx\\ &\quad\frac{dy}{dx}=\frac{b^2 x}{a^2 y} \end{align*} \begin{align*} \text{Tangent at $P$:}&\quad\bf\frac{x_1 x}{a^2}-\frac{y_1 y}{b^2}=1\\ &\quad y-y_1=\left(\frac{b^2 x_1}{a^2 y_1}\right)\left(x-x_1\right)\\ \times\frac{y_1}{b^2}:&\quad\frac{y_1}{b^2}y-\frac{y_1}{b^2}y_1=\frac{x_1}{a^2}x-\frac{x_1}{a^2}x_1\\ &\quad\frac{x_1 x}{a^2}-\frac{y_1 y}{b^2}=\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}=1 \end{align*} \begin{align*} \text{Normal at $P$:}&\quad\bf\frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2\\ &\quad\text{Gradient is}\quad -\frac{a^2}{b^2}\frac{y_1}{x_1}\\ &\quad y-y_1=\left(-\frac{a^2}{b^2}\frac{y_1}{x_1}\right)\left(x-x_1\right)\\ \times\frac{b^2}{y_1}:&\quad\frac{b^2}{y_1}y-\frac{b^2}{\not{y_1}}\not{y_1}=-\frac{a^2}{x_1}x+\frac{a^2}{\not{x_1}}\not{x_1}\\ &\quad\frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2 \end{align*} \begin{align*} \text{Intersection $T$:}&\quad\bf(x_0,y_0)=\left(\frac{a^2(y_2-y_1)}{x_1 y_2-x_2 y_1},\frac{b^2(x_2-x_1)}{x_1 y_2-x_2 y_1}\right)\\ PT:&\quad\frac{x_1 x}{a^2}-\frac{y_1 y}{b^2}=1\tag{1}\\ QT:&\quad\frac{x_2 x}{a^2}-\frac{y_2 y}{b^2}=1\tag{2}\\ (1)\times y_2-(2)\times y_1:&\quad\frac{x_1 y_2 x}{a^2}-\frac{x_2 y_1 x}{a^2}=y_2-y_1\\ &\quad(x_1 y_2-x_2 y_1)x=a^2(y_2-y_1)\\ &\quad x=\frac{a^2(y_2-y_1)}{x_1 y_2-x_2 y_1}\\ (1)\times x_2-(2)\times x_1:&\quad -\frac{x_2 y_1 y}{b^2}+\frac{x_1 y_2 y}{b^2}=x_2-x_1\\ &\quad(x_1 y_2-x_2 y_1)y=b^2(x_2-x_1)\\ &\quad y=\frac{b^2(x_2-x_1)}{x_1 y_2-x_2 y_1} \end{align*} \begin{align*} \text{Chord of Contact $PQ$:}&\quad\bf\frac{x_0 x}{a^2}-\frac{y_0 y}{b^2}=1\\ PT:&\quad\frac{x_1 x}{a^2}-\frac{y_1 y}{b^2}=1\\ QT:&\quad\frac{x_2 x}{a^2}-\frac{y_2 y}{b^2}=1\\ &\quad\text{$T(x_0,y_0)$ is on both $PT$ and $QT$, so}\\ &\quad\frac{x_1 x_0}{a^2}-\frac{y_1 y_0}{b^2}=1\\ &\quad\frac{x_2 x_0}{a^2}-\frac{y_2 y_0}{b^2}=1\\ &\quad\frac{x_0 x}{a^2}-\frac{y_0 y}{b^2}=1\:\:\text{satisfies both $P$ and $Q$,}\\ &\qquad\qquad\qquad\text{and therefore is the Chord of Contact.} \end{align*} % % Hyperbola (Parametric) % \begin{align*} \text{{\bf Hyperbola} (Parametric)}\qquad x=a\sec\theta,&\:\:y=b\tan\theta\qquad\qquad\qquad\qquad\qquad\qquad \end{align*} \begin{align*} \text{Derivative:}&\quad\bf\frac{b\sec\theta}{a\tan\theta}\\ &\quad\frac{dx}{d\theta}=a\sec\theta\tan\theta\:,\quad\frac{dy}{d\theta}=b\sec^2\theta\\ &\quad\frac{dy}{dx}=\frac{b\sec\theta}{a\tan\theta} \end{align*} \begin{align*} \text{Tangent at $P$:}&\quad\bf\frac{x\sec\theta_1}{a}-\frac{y\tan\theta_1}{b}=1\\ &\quad y-y_1=\frac{b\sec\theta_1}{a\tan\theta_1}(x-x_1)\\ \times\frac{\tan\theta_1}{b}:&\quad \frac{\tan\theta_1}{b}(y-b\tan\theta_1)=\frac{\sec\theta_1}{a}(x-a\sec\theta_1)\\ &\quad \frac{y\tan\theta_1}{b}-\tan^2\theta_1=\frac{x\sec\theta_1}{a}-\sec^2\theta_1\\ &\quad \frac{x\sec\theta_1}{a}-\frac{y\tan\theta_1}{b}=\sec^2\theta_1-\tan^2\theta_1=1 \end{align*} \begin{align*} \text{Normal at $P$:}&\quad\bf\frac{ax}{\sec\theta_1}+\frac{by}{\tan\theta_1}=a^2+b^2\\ &\quad y-y_1=-\frac{a\tan\theta_1}{b\sec\theta_1}(x-x_1)\\ \times\frac{b}{\tan\theta_1}:&\quad \frac{b}{\tan\theta_1}(y-b\tan\theta_1)=-\frac{a}{\sec\theta_1}(x-a\sec\theta_1)\\ &\quad \frac{by}{\tan\theta_1}-b^2=-\frac{ax}{\sec\theta_1}+a^2\\ &\quad \frac{ax}{\sec\theta_1}+\frac{by}{\tan\theta_1}=a^2+b^2 \end{align*} \begin{align*} \text{Intersection $T$:}&\quad\bf\left(\frac{a\sin(\theta_1-\theta_2)}{\sin\theta_1-\sin\theta_2},\:\frac{-b(\cos\theta_1-\cos\theta_2)}{\sin\theta_1-\sin\theta_2}\right)\\ PT:&\quad\frac{x\sec\theta_1}{a}-\frac{y\tan\theta_1}{b}=1\tag{1}\\ QT:&\quad\frac{x\sec\theta_2}{a}-\frac{y\tan\theta_2}{b}=1\tag{2}\\ (1)\times\tan\theta_2-(2)\times\tan\theta_1:&\quad\frac{\:x\:}{a}(\sec\theta_1\tan\theta_2-\tan\theta_1\sec\theta_2)=\tan\theta_2-\tan\theta_1\\ &\quad\frac{\:x\:}{a}\left(\frac{1}{\cos\theta_1}\cdot\frac{\sin\theta_2}{\cos\theta_2}-\frac{\sin\theta_1}{\cos\theta_1}\cdot\frac{1}{\cos\theta_2}\right)=\frac{\sin\theta_2}{\cos\theta_2}-\frac{\sin\theta_1}{\cos\theta_1}\\ &\quad\frac{\:x\:}{a}\left(\frac{\sin\theta_2-\sin\theta_1}{\cos\theta_1\cos\theta_2}\right)=\frac{\sin\theta_2\cos\theta_1-\cos\theta_2\sin\theta_1}{\cos\theta_1\cos\theta_2}=\frac{\sin(\theta_2-\theta_1)}{\cos\theta_1\cos\theta_2}\\ &\quad x=\frac{a\sin(\theta_1-\theta_2)}{\sin\theta_1-\sin\theta_2}\\ (1)\times\sec\theta_2-(2)\times\sec\theta_1:&\quad-\frac{\:y\:}{b}(\sec\theta_2\tan\theta_1-\tan\theta_2\sec\theta_1)=\sec\theta_2-\sec\theta_1\\ &\quad-\frac{\:y\:}{b}\left(\frac{1}{\cos\theta_2}\cdot\frac{\sin\theta_1}{\cos\theta_1}-\frac{\sin\theta_2}{\cos\theta_2}\cdot\frac{1}{\cos\theta_1}\right)=\frac{1}{\cos\theta_2}-\frac{1}{\cos\theta_1}\\ &\quad-\frac{\:y\:}{b}\left(\frac{\sin\theta_1-\sin\theta_2}{\cos\theta_2\cos\theta_1}\right)=\frac{\cos\theta_1-\cos\theta_2}{\cos\theta_2\cos\theta_1}\\ &\quad y=\frac{-b(\cos\theta_1-\cos\theta_2)}{\sin\theta_1-\sin\theta_2}\\ \end{align*} \begin{align*} \text{Chord of Contact $PQ$:}&\quad\bf\frac{\:x\:}{a}\cos\left(\tfrac{\theta_1-\theta_2}{2}\right)-\frac{\:y\:}{b}\sin\left(\tfrac{\theta_1+\theta_2}{2}\right)=\cos\left(\tfrac{\theta_1+\theta_2}{2}\right)\\ \text{First Prove:}&\quad\frac{x}{a}\cdot\frac{\sin(\theta_1-\theta_2)}{\sin\theta_1-\sin\theta_2}+\frac{y}{b}\cdot\frac{\cos\theta_1-\cos\theta_2}{\sin\theta_1-\sin\theta_2}=1\\ PQ:&\quad\frac{y-b\tan\theta_1}{x-a\sec\theta_1}=\frac{b\tan\theta_1-b\tan\theta_2}{a\sec\theta_1-a\sec\theta_2}\\ &\quad (y-b\tan\theta_1)\cdot(a\sec\theta_1-a\sec\theta_2)=(x-a\sec\theta_1)\cdot(b\tan\theta_1-b\tan\theta_2)\\ &\quad ay(\sec\theta_1-\sec\theta_2)-ab\tan\theta_1(\sec\theta_1-\sec\theta_2)=bx(\tan\theta_1-\tan\theta_2)-ab\sec\theta_1(\tan\theta_1-\tan\theta_2)\\ &\quad ay(\sec\theta_1-\sec\theta_2)-bx(\tan\theta_1-\tan\theta_2)=ab\tan\theta_1(\sec\theta_1-\sec\theta_2)-ab\sec\theta_1(\tan\theta_1-\tan\theta_2)\\ \times\frac{1}{ab}:&\quad \frac{y}{b}(\sec\theta_1-\sec\theta_2)-\frac{x}{a}(\tan\theta_1-\tan\theta_2)=\tan\theta_1(\sec\theta_1-\sec\theta_2)-\sec\theta_1(\tan\theta_1-\tan\theta_2)\\ &\quad \frac{y}{b}(\sec\theta_1-\sec\theta_2)-\frac{x}{a}(\tan\theta_1-\tan\theta_2)=-\tan\theta_1\sec\theta_2+\sec\theta_1\tan\theta_2\\ &\quad\frac{y}{b}\left(\frac{1}{\cos\theta_1}-\frac{1}{\cos\theta_2}\right)-\frac{x}{a}\left(\frac{\sin\theta_1}{\cos\theta_1}-\frac{\sin\theta_2}{\cos\theta_2}\right)=-\frac{\sin\theta_1}{\cos\theta_1}\cdot\frac{1}{\cos\theta_2}+\frac{1}{\cos\theta_1}\cdot\frac{\sin\theta_2}{\cos\theta_2}\\ &\quad\frac{y}{b}\left(\frac{\cos\theta_2-\cos\theta_1}{\cos\theta_1\cos\theta_2}\right)-\frac{x}{a}\left(\frac{\sin\theta_1\cos\theta_2-\cos\theta_1\sin\theta_2}{\cos\theta_1\cos\theta_2}\right)=\frac{\sin\theta_2-\sin\theta_1}{\cos\theta_1\cos\theta_2}\\ &\quad\frac{x}{a}\left(\frac{\sin(\theta_1-\theta_2)}{\cos\theta_1\cos\theta_2}\right)-\frac{y}{b}\left(\frac{\cos\theta_2-\cos\theta_1}{\cos\theta_1\cos\theta_2}\right)=\frac{\sin\theta_1-\sin\theta_2}{\cos\theta_1\cos\theta_2}\\ \therefore&\quad\frac{x}{a}\cdot\frac{\sin(\theta_1-\theta_2)}{\sin\theta_1-\sin\theta_2}+\frac{y}{b}\cdot\frac{\cos\theta_1-\cos\theta_2}{\sin\theta_1-\sin\theta_2}=1\\ &\quad\frac{\:x\:}{a}\cdot\frac{\not{2}\not{\sin\left(\frac{\theta_1-\theta_2}{2}\right)}\cos\left(\frac{\theta_1-\theta_2}{2}\right)}{\not{2}\cos\left(\frac{\theta_1+\theta_2}{2}\right)\not{\sin\left(\frac{\theta_1-\theta_2}{2}\right)}}+\frac{\:y\:}{b}\cdot\frac{-\not{2}\sin\left(\frac{\theta_1+\theta_2}{2}\right)\not{\sin\left(\frac{\theta_1-\theta_2}{2}\right)}}{\not{2}\cos\left(\frac{\theta_1+\theta_2}{2}\right)\not{\sin\left(\frac{\theta_1-\theta_2}{2}\right)}}=1\\ &\quad\frac{\:x\:}{a}\cdot\frac{\cos\left(\frac{\theta_1-\theta_2}{2}\right)}{\cos\left(\frac{\theta_1+\theta_2}{2}\right)}-\frac{\:y\:}{b}\cdot\frac{\sin\left(\frac{\theta_1+\theta_2}{2}\right)}{\cos\left(\frac{\theta_1+\theta_2}{2}\right)}=1\\ \therefore&\quad\frac{\:x\:}{a}\cos\left(\tfrac{\theta_1-\theta_2}{2}\right)-\frac{\:y\:}{b}\sin\left(\tfrac{\theta_1+\theta_2}{2}\right)=\cos\left(\tfrac{\theta_1+\theta_2}{2}\right) \end{align*} \end{document}