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Rectangular Hyperbola.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=0.5cm]{geometry} \begin{document} \noindent {\Large Rectangular Hyperbola} \begin{align*} &\text{A rectangular hyperbola is one with both asymptotes meet at right angles. i.e. $y=\pm x$.}\\ &\because\quad\text{The asymptotes of a hyperbola is $\frac{\:x\:}{a}=\pm\frac{\:y\:}{b}$,}\qquad \therefore\quad a=b\:,\text{ hence the following properties:} \end{align*} % % Basics % \begin{align*} \text{When foci are on x-axis:}\\ \text{Cartesian equation:}&\quad\frac{x^2}{a^2}-\frac{y^2}{a^2}=1\qquad\text{or}\qquad\boxed{x^2-y^2=a^2}\\ \text{Eccentricity:}&\quad e=\sqrt{1+\frac{a^2}{a^2}}=\sqrt{2}\\ \text{Foci:}&\quad S=(a\sqrt{2},0),\quad S'=(-a\sqrt{2},0)\\ \text{Directrices:}&\quad m:\:x=\frac{\:a\:}{\sqrt{2}},\quad m':\:x=-\frac{\:a\:}{\sqrt{2}}\\ \text{Asymptotes:}&\quad x=\pm y\\ &\ldots\text{ similar to those properties of a hyperbola in general.}\\ % \text{When the above hyperbola}&\text{ is rotated anti-clockwise by $\tfrac{\pi}{4}$ on a complex plan:}\\ % \text{Before (foci on x-axis):}\quad&|z-ae|=e\left[Re(z)-\frac{a}{e}\right]=e\left(\frac{z+\overline{z}}{2}-\frac{a}{e}\right)\:,\quad\left|z-a\sqrt{2}\right|=\sqrt{2}\left(\frac{z+\overline{z}}{2}-\frac{a}{\sqrt{2}}\right)\\ &\left|z-a\sqrt{2}\right|^2=2\left(\frac{z+\overline{z}}{2}-\frac{a}{\sqrt{2}}\right)^2\:, \quad 2\:\left(z-a\sqrt{2}\right)\cdot\left(\overline{z}-a\sqrt{2}\right)-4\left(\frac{z+\overline{z}}{2}-\frac{a}{\sqrt{2}}\right)^2=0\\ &2\:\left(z\overline{z}-a\sqrt{2}z-a\sqrt{2}\overline{z}+2a^2\right)-\left(z+\overline{z}-a\sqrt{2}\right)^2=0\\ &\left(2z\overline{z}-2za\sqrt{2}-2\overline{z}a\sqrt{2}+4a^2\right)-\left(z^2+\overline{z}^2+2a^2+2z\overline{z}-2za\sqrt{2}-2\overline{z}a\sqrt{2}\right)=0\\ &2a^2-z^2-\overline{z}^2=0\:,\quad\boxed{z^2+\overline{z}^2=2a^2}\\ % \text{After (foci on $y=x$):}\quad&\text{Rotate clockwise by $\tfrac{\pi}{4}$ will fit the above. i.e. to substitute $z$ by }(\cos\tfrac{\pi}{4}-i\sin\tfrac{\pi}{4})z=\tfrac{1}{\sqrt{2}}(1-i)z\:.\\ &\left[\tfrac{1}{\sqrt{2}}(1-i)z\right]^2+\overline{\left[\tfrac{1}{\sqrt{2}}(1-i)z\right]}^2=2\:a^2\\ &\left[(1-i)(x+iy)\right]^2+\left[(1+i)(x-iy)\right]^2=4\:a^2\\ &\left[x+y-i(x-y)\right]^2+\left[x+y+i(x-y)\right]^2=4\:a^2\\ &2\left[(x+y)^2-(x-y)^2\right]=4\:a^2\:,\quad 8\:xy=4\:a^2\:,\quad\boxed{xy=\frac{\:\:a^2}{2}}\\ % &\text{Since the foci are on $y=x$, the vertices are }\left(\pm\frac{a}{\sqrt{2}},\pm\frac{a}{\sqrt{2}}\right)\text{ or }(\pm c,\pm c),\text{ where }\boxed{c=\frac{a}{\sqrt{2}}}\:.\\ &\text{The Rectangular Hyperbola becomes }\boxed{xy=c^2}\:.\\ % \text{Foci rotated by $\tfrac{\pi}{4}$:}\quad&\text{When the foci is on x-axis, they are }\pm a\sqrt{2}\\ &\text{When the foci is on $y=x$, they are rotated by $\tfrac{\pi}{4}$: }\pm\tfrac{1}{\sqrt{2}}(1+i)\cdot a\sqrt{2}=\pm a\pm ia\\ % \text{Foci:}\quad&\boxed{S=(a,a),\quad S'=(-a,-a)}\quad\text{or}\quad\boxed{S=(c\sqrt{2},c\sqrt{2}),\quad S'=(-c\sqrt{2},-c\sqrt{2})}\\ &\text{The directrices of a ``horizontal'' hyperbola on complex plan: }Re(z)=\pm\frac{\:a\:}{\sqrt{2}}\\ &\text{The directrices after rotated by $-\tfrac{\pi}{4}$ will become }Re(\tfrac{1}{\sqrt{2}}(1-i)z)=\pm\frac{\:a\:}{\sqrt{2}}\\ &Re[(1-i)(x+iy)]=\pm a\:,\quad Re[(x+y)-i(x-y)]=\pm a\:,\quad x+y=\pm a\\ % \text{Directrices:}&\quad\boxed{x+y=\pm a}\quad\text{or}\quad\boxed{x+y=\pm c\sqrt{2}}\\ % \text{Asymptotes:}&\quad\boxed{x=0\:,\quad y=0}\qquad\text{($x=\pm y$ rotated by $\tfrac{\pi}{4}$)}\\ \end{align*} % % % \begin{align*} \text{\large{\bf Tangents and Chords:}}&\quad\text{In the following,}\\ &\quad\text{$P(x_1,y_1)$ or $P(cp,\tfrac{c}{p})$ is on the hyperbola, so is}\\ &\quad\text{$Q(x_2,y_2)$ or $Q(cq,\tfrac{c}{q})$, and}\\ &\quad\text{$T(x_0,y_0)$ or $T(ct,\tfrac{c}{t})$ is the intersection of the}\\ &\qquad\qquad\text{two tangents from $P$ and $Q$.}\\ \\ % % Tangents and Chords (Cartesian) % \text{\large Cartesian Form:}&\quad xy=c^2\\ % \text{Derivative:}&\quad y\:dx+x\:dy=0\:,\quad\frac{dy}{dx}=-\frac{\:y\:}{x}\\ \text{Tangent at $P$:}&\quad\frac{y-y_1}{x-x_1}=-\frac{y_1}{x_1}\:,\quad x_1 y-x_1 y_1+xy_1-x_1 y_1 = 0\:,\quad\text{(Note: $x_1 y_1=c^2$)}\\ &\quad\boxed{y_1 x+x_1 y=2c^2}\\ \text{Normal at $P$:}&\quad\frac{y-y_1}{x-x_1}=\frac{x_1}{y_1}\:,\quad y_1 y-y_1^2-x_1 x+x_1^2=0\:, \quad\boxed{x_1 x-y_1 y=x_1^2-y_1^2}\\ \text{Intersection $T$:}&\quad\boxed{(x_0,y_0)=\left[\frac{2c^2(x_1-x_2)}{x_1 y_2-x_2 y_1},\frac{-2c^2(y_1-y_2)}{x_1 y_2-x_2 y_1}\right]}\\ \text{Chord of Contact $PQ$:}&\quad\boxed{y_0 x+x_0 y=2c^2}\quad\text{as $(x_0,y_0)$ is on both}\quad yx_1+xy_1=2c^2\quad\text{and}\quad yx_2+xy_2=2c^2\\ \text{So}&\quad\frac{-2c^2(y_1-y_2)}{x_1 y_2-x_2 y_1}x+\frac{2c^2(x_1-x_2)}{x_1 y_2-x_2 y_1}y=2c^2\\ &\quad (y_2-y_1)x+(x_1-x_2)y=x_1 y_2-x_2 y_1\\ &\quad\left(\frac{c^2}{x_2}-\frac{c^2}{x_1}\right)x+(x_1-x_2)y=x_1\frac{c^2}{x_2}-x_2\frac{c^2}{x_1}\\ \times x_1 x_2:&\quad c^2(x_1-x_2)x+x_1 x_2(x_1-x_2)y=c^2(x_1^2-x_2^2)=c^2(x_1-x_2)(x_1+x_2)\\ &\quad\boxed{c^2 x+x_1 x_2 y=c^2(x_1+x_2)}\\ \\ % % Tangents and Chords (Parametric) % \text{\large Parametric Form:}&\quad x=ct,\:\:y=\frac{\:c\:}{t}\:,\quad\text{where }t\neq 0\\ % \text{Derivative:}&\quad\frac{dy}{dx}=\frac{-ct^{-2}}{c}=-\frac{\:1}{\:t^2}\\ \text{Tangent at $P$:}&\quad\frac{y-\frac{c}{p}}{x-cp}=-\frac{1}{p^2}\:,\quad\boxed{x+p^2 y-2cp=0}\\ \text{Normal at $P$:}&\quad\frac{y-\frac{c}{p}}{x-cp}=p^2\:,\quad p^2(x-cp)-(y-\frac{c}{p})=0\:,\\ \times\frac{\:1\:}{p}:&\quad px-cp^2-\frac{\:y\:}{p}+\frac{c}{p^2}=0\:,\quad\boxed{px-\frac{\:1\:}{p}y=c\left(p^2-\frac{\:1\:}{p^2}\right)}\\ \text{Intersection $T$:}&\quad(x_0,y_0) =\left[\frac{2c^2(cp-cq)}{cp\cdot\frac{c}{q}-cq\cdot\frac{c}{p}}\:,\frac{-2c^2\left(\frac{c}{p}-\frac{c}{q}\right)}{cp\cdot\frac{c}{q}-cq\cdot\frac{c}{p}}\right]\\ &\quad=\left[\frac{2c(p-q)}{\frac{p}{q}-\frac{q}{p}}\:,\frac{-2c\left(\frac{q-p}{pq}\right)}{\frac{p}{q}-\frac{q}{p}}\right] =\left[\frac{2cpq(p-q)}{p^2-q^2}\:,\frac{-2c(q-p)}{p^2-q^2}\right]\\ &\quad=\boxed{\left(\frac{2cpq}{p+q}\:,\frac{2c}{p+q}\right)}\\ \text{Chord of Contact $PQ$:}&\quad\frac{2cx}{p+q}+\frac{2cpqy}{p+q}=2c^2\\ &\quad\boxed{x+pqy=c(p+q)}\\ \end{align*} \end{document}