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Parametric representations

  1. Aug 27, 2007 #1
    Sketch and represent parametrically the following: (a) [tex] \mid z+a+\iota b\mid =r \ \mbox { clockwise}\\ [/tex], (b) ellipse [tex] 4(x-1)^2 + 9(y+2)^2 =36 \ [/tex].
    Taking (a) first [tex] \mid z + a + \iota b \mid = r \mbox{- is the distance between the complex numbers }\ z=x+\iota y \ \mbox{ and } \ a + \iota b \ \mbox{ if the distance is always r, then we have } \ \sqrt{(x+a)^2 + (y+b)^2} = r \\[/tex]. This is a circle with center -a -ib and radius r, But the circle at the origin can be parametrically represented as [tex] r\exp{\iota t} \ 0 \leq \ t \ \leq 2\pi \\[/tex] but since t goes from [tex] 0 \mbox{ to } 2\pi \\[/tex], clockwise it's equation is [tex] \exp{-\iota t} \\ [/tex], therefore the circle is [tex] -a -\iota b + r\exp{-\iota t} = 0 \\[/tex]
    Am I correct with (a)'s reasoning. I don't know how to do (b) as I know nothing about an ellipse. Thanks for the help.
  2. jcsd
  3. Aug 27, 2007 #2
    To get you started on b):

    The equation for an ellipse centered at (0,0) is:

    [tex]\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1[/tex]

    Which has the parametric representation

  4. Aug 27, 2007 #3
    For part (a) you are on the right track. Squaring both sides we get the equation for a circle with center (-a,-b) in the plane. Take the parametrization of a circle at the origin and perform horizontal and vertical shifts to it to get the parametrization of this particular circle.

    The last equation is not correct. You should have the LHS in abs, and on the RHS there should be an r. squaring this equation, we just get the standard equation of circle.
  5. Aug 27, 2007 #4
    Thanks for the reply
  6. Aug 28, 2007 #5
    For the ellipse [tex] 4(x-1)^2 + 9(y+2)^2 = 36 \mbox{. Therefore } \frac{(x-1)^2}{9} + \frac{(y+2)^2}{4} = 1 \\ = \frac{(x-1)^2}{3^2} + \frac{(y+2)^2}{2^2} [/tex]. This is an ellipse with semi-major axis and semi-minor axis of lengths 3 and 2 respectively whose center is at (1,-2). Therefore its equation is [tex] \mid 4x + 9 \iota y -1 +2 \iota\mid + 2\cos\theta +3\iota\sin\theta = 6 \ \mbox{ for} 0 \leq \theta \ \leq 2\pi \\[/tex].
    About (a) do you mean [tex] \mid -a-\iota b +r\exp{-\iota t\mid =r [/tex]Thanks for helping.
    Last edited: Aug 28, 2007
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