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BBy Bot
Nov 03'24

Exercise

[math] \newcommand{\ex}[1]{\item } \newcommand{\sx}{\item} \newcommand{\x}{\sx} \newcommand{\sxlab}[1]{} \newcommand{\xlab}{\sxlab} \newcommand{\prov}[1] {\quad #1} \newcommand{\provx}[1] {\quad \mbox{#1}} \newcommand{\intext}[1]{\quad \mbox{#1} \quad} \newcommand{\R}{\mathrm{\bf R}} \newcommand{\Q}{\mathrm{\bf Q}} \newcommand{\Z}{\mathrm{\bf Z}} \newcommand{\C}{\mathrm{\bf C}} \newcommand{\dt}{\textbf} \newcommand{\goesto}{\rightarrow} \newcommand{\ddxof}[1]{\frac{d #1}{d x}} \newcommand{\ddx}{\frac{d}{dx}} \newcommand{\ddt}{\frac{d}{dt}} \newcommand{\dydx}{\ddxof y} \newcommand{\nxder}[3]{\frac{d^{#1}{#2}}{d{#3}^{#1}}} \newcommand{\deriv}[2]{\frac{d^{#1}{#2}}{dx^{#1}}} \newcommand{\dist}{\mathrm{distance}} \newcommand{\arccot}{\mathrm{arccot\:}} \newcommand{\arccsc}{\mathrm{arccsc\:}} \newcommand{\arcsec}{\mathrm{arcsec\:}} \newcommand{\arctanh}{\mathrm{arctanh\:}} \newcommand{\arcsinh}{\mathrm{arcsinh\:}} \newcommand{\arccosh}{\mathrm{arccosh\:}} \newcommand{\sech}{\mathrm{sech\:}} \newcommand{\csch}{\mathrm{csch\:}} \newcommand{\conj}[1]{\overline{#1}} \newcommand{\mathds}{\mathbb} [/math]

It has been shown that the distance between a point on the ellipse

[[math]] \frac{x^2}{a^2} + \frac{y^2}{a^2-c^2} = 1 [[/math]]

and the focus [math](c,0)[/math] is [math]\left| \frac{xc}a - a \right|[/math].

  • Show that this distance is [math]a - \frac{xc}a[/math] for [math]|x| \leq a[/math].
  • Show that the distance between a point on the ellipse and the focus [math](-c,0)[/math] is [math]\left| \frac{xc}a + a \right|[/math] and that the distance is [math]a + \frac{xc}a[/math].
  • Show that the sum of the distances from a point on the ellipse to the foci is [math]2a[/math] and hence that the graph of [math]\frac{x^2}{a^2} + \frac{y^2}{a^2-c^2} = 1[/math] contains only those points which satisfy the locus definition.