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2015-02-17 Ellipse Tangential Circle.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage{graphicx} \usepackage{cancel} \usepackage{wrapfig} \usepackage{txfonts} \usepackage[hmargin=2cm,vmargin=2cm]{geometry} \parindent0em \setlength{\parskip}{0.3cm} \begin{document} %\hspace*{-0.5cm}{\large Cambridge 4U p. 103 Q4} \hspace*{-0.5cm}{\bf Y12 4U 2005 Q3 a iii.} (a) Ellipse $E:\quad\dfrac{x^2}{100}+\dfrac{y^2}{75}=1$. (iii) A circle is tangential to the ellipse $E$ at $P(5,7\frac{1}{2})$ and at $Q(5,-7\frac{1}{2})$. Show that the centre of the circle is $(1\frac{1}{4},0)$. \hspace*{-0.5cm}{\bf Solution:} (Note: This solution is an alternative to using the result of (ii), the normals.) Strategy: Let the centre of the circle be $(h,0)$. %The radius is therefore\quad$r=\sqrt{(5-h)^2+(7\frac{1}{2})^2}$. %=(5-h)^2+(7\frac{1}{2})^2 %$x^2-2hx+\cancel{h^2}+y^2=25-10h+\cancel{h^2}+\frac{225}{4}$,\\ The equation of the circle $C$ is\quad$(x-h)^2+y^2=r^2$. Differentiate both sides:\quad $d[(x-h)^2+y^2]=2(x-h)~dx+2y~dy=0,$\qquad $\dfrac{dy}{dx}=-\dfrac{x-h}{y}.$ The gradient of tangent at $P(5,7\frac{1}{2})$: For $E$:\quad$m_E=-\frac{b^2x}{a^2y}=-\dfrac{75\times 5}{100\times 7\frac{1}{2}}=-\dfrac{1}{2}.$ For $C$:\quad$m_C=-\dfrac{x-h}{y}=\dfrac{h-5}{7\frac{1}{2}}=\dfrac{2h-10}{15}.$ $\because~~m_C=m_E,\quad\therefore~~\dfrac{2h-10}{15}=-\dfrac{1}{2},\quad h=1\frac{1}{4}.$ The centre of the circle $(h,0)=(1\frac{1}{4},0)$. \end{document}