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Sagot :
The polar equation of an ellipse is [tex]r=-\frac{3}{1+2cos\theta}[/tex].
The vertices of ellipse are (−1,0) and (3,0).
The polar equation of an ellipse can be represented as
[tex]r=\frac{ep}{1+ecos\theta}[/tex]
where e is the eccentricity.
Eccentricity, e = [tex]\frac{c}{a}[/tex]
c is the distance from the center to the focus and a is the distance from the center to the vertex
[tex]c=\frac{3-(-1)}{2}[/tex]
⇒ [tex]c=\frac{4}{2}[/tex]
⇒ c = 2
[tex]a=\frac{3+(-1)}{2}[/tex]
⇒ [tex]a=\frac{2}{2}[/tex]
⇒ a = 1
Then, e = [tex]\frac{2}{1}[/tex]
⇒ e = 2
Now, the polar equation of an ellipse becomes as,
⇒ [tex]r=\frac{2p}{1+2cos\theta}[/tex] ------- (1)
Now plug in a vertex point such as (-1,0) and solve for p,
⇒ [tex]-1=\frac{2p}{1+2cos0}[/tex]
⇒ [tex]-1=\frac{2p}{1+2(1)}[/tex] [∵ cos 0 = 1]
⇒ [tex]-1=\frac{2p}{3}[/tex]
⇒ [tex]-3=2p[/tex]
⇒ [tex]p=-\frac{3}{2}[/tex]
Thus the polar equation of an ellipse (1) becomes as,
⇒ [tex]r=\frac{2(-\frac{3}{2} )}{1+2cos\theta}[/tex]
⇒ [tex]r=-\frac{3}{1+2cos\theta}[/tex]
Hence we can conclude that the polar equation of an ellipse is [tex]r=-\frac{3}{1+2cos\theta}[/tex].
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