MATHHX A
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3.7 Ellipser
En ellipse er en en cirkel der er ”strukket ud”. Her er et eksempel:
Figur 3.1: Ellipse med centrum i \(C(x_0,y_0)\) og halvakser \(a\) og \(b\) .
Ovenstående figur viser en ellipse med centrum i \(C(x_0,y_0)\). Længderne af stykket \(a\) kaldes ellipsens vandrette halvakser og længden af stykket \(b\) kaldes ellipsens lodrette halvakse. Den største af de to halvakser kaldes
også den halve storakse mens den mindste kaldes ellipsens halve lilleakse . Læg mærke til at \(a\) altid måles vandret og \(b\) lodret.
Øvelse 3.7.1
Betragt ellipsen:
a) Bestem den vandrette halvakse, den lodrette halvakse og centrum for ellipsen.
b) Bestem den halve storakse og den halve lilleakse.
c) Bestem \(a\), \(b\), \(x_0\) og \(y_0\) for ellipsen.
Løsning 3.7.1
a) Den vandrette halvakse er \(1\), den lodrette halvakse er \(2\) og centrum er i punktet \((4,2)\).
b) Den halve storakse er \(2\), den halve lilleakse er \(1\).
c) \(a=1\), \(b=2\), \(x_0=4\) og \(y_0=2\)
Ligesom vi har cirklens ligning har vi også ellipsens ligning:
Sætning 3.7.1
En ellipse med centrum i \(C=(x_0,y_0)\) og halvakser \(a\) og \(b\) har ligningen
\(\seteqnumber{0}{3.}{1}\)
\begin{equation}
\label {eliplig1} \frac {(x-x_0)^2}{a^2}+\frac {(y-y_0)^2}{b^2}=1
\end{equation}
Øvelse 3.7.2
Angiv centrum og halvakser for følgende ellipserne givet ved ligningerne:
a) \(\frac {(x-9)^2}{25}+\frac {(y-14)^2}{16}=1\)
b) \(\frac {(x-3)^2}{9}+\frac {(y+2)^2}{36}=1\)
c) \(\frac {(x-5)^2}{5}+\frac {y^2}{81}=1\)
d) \(x^2 + \frac {y^2}{4}=1\)
Løsning 3.7.2
a) Centrum er \((9,14)\), den vandrette halvakse er \(5\), den lodrette halvakse er \(4\).
b) Centrum er \((3,-2)\), den vandrette halvakse er \(3\), den lodrette halvakse er \(6\).
c) Centrum er \((5,0)\), den vandrette halvakse er \(\sqrt {5}\), den lodrette halvakse er \(9\).
d) Centrum er \((0,0)\), den vandrette halvakse er \(1\), den lodrette halvakse er \(2\).
Øvelse 3.7.3
En ellipse har vandret halvakse \(4\), lodret halvakse \(2\) og centrum i \((-2,1)\)
a) Opskriv en ligning for ellipsen.
b) Skriv ligningen på formen \(ax^2+bx+cy^2+dy+e=0\).
c) Tjek ved beregning om punktet \((-2,1)\) ligger på ellipsen.
Øvelse 3.7.4
Betragt ellipsen med ligningen
\[\frac {(x-5)^2}{9}+\frac {(y-4)^2}{4}=1\]
Øvelse 3.7.5
Antag at vi har en ellipse, hvor de to halvakser er ens.
Eksempel 3.7.1
Vi vil vise at ligningen \(4x^2+y^2-16x+6y=-9\) er ligning for en ellipse. Vi starter med at samle \(x\)’erne og \(y\)’erne:
\[4x^2-16x+y^2+6y=-9.\]
Nu faktorisere vi leddene med \(x\) (havde der stået en konstant foran \(y^2\), skulle leddene med \(y\) også faktoriseres):
\[4(x^2-4x)+y^2+6y=-9.\]
Så kvadratkompletterer vi
\[4((x-2)^2-4)+(y+3)^2-9=-9.\]
Så ganges parentesen ud:
\[4(x-2)^2-16+(y+3)^2-9=-9,\]
og vi samler konstanterne på højresiden:
\[4(x-2)^2+(y+3)^2=-9+16+9,\]
og regner højresiden:
\[4(x-2)^2+(y+3)^2=16.\]
Vi kan se på ellipsens ligning (3.2 ) at højresiden skal være \(1\), så vi dividerer med \(16\) på begge sider:
\[\frac {4(x-2)^2}{16}+\frac {(y+3)^2}{16}=1.\]
Vi forkorter:
\[\frac {(x-2)^2}{4}+\frac {(y+3)^2}{16}=1,\]
og skriver nævnerne som et kvadrat (noget i anden):
\[\frac {(x-2)^2}{2^2}+\frac {(y+3)^2}{4^2}=1.\]
Ved sammenligning med ellipsens ligning (3.2 ) ses at vi har en ellipse med centrum i \((2,-3)\) og lille halvakse \(2\) og store halvakse \(4\). Puha det var hårdt arbejde, godt det ikke
er mig som skal regne de efterfølgende øvelser.
Øvelse 3.7.7
Bestem ellipsens centrum og halvakser. Brug metoden fra ovenstående eksempel.
a) \(x^2 + 4y^2 - 6x - 8y = 3\)
b) \(25x^2 + 4y^2 + 50x - 24y = 39\)
c) \(9x^2 + y^2 + 36x + 14y + 76 = 0\)
d) \(8x^2 + 32 y^2 - 32x = 0\) (VINK: du kan forsimple den først, ved at dividere igennem med et passende tal)
Løsning 3.7.7
a) Centrum er i \((3,1)\), vandret halvakse er \(4\) og lodret halvakse er \(2\).
b) Centrum er i \((-1,3)\), vandret halvakse er \(2\) og lodret halvakse er \(5\).
c) Centrum er i \((-2,-7)\), vandret halvakse er \(1\) og lodret halvakse er \(3\).
d) Centrum er i \((2,0)\), vandret halvakse er \(2\) og lodret halvakse er \(1\).