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Analytic geometry

  1. May 31, 2005 #1
    Hi all. I have a analytic geometry question that I need a bit of help with.
    consider the concentric circles with the equations:

    [tex]x^2 + y^2 = 9[/tex]
    [tex]x^2 +y^2 = 4[/tex]

    A radius from the center O intersects the inner circle at P and the outer circle at Q. The line paralell to the x-axis through P meets the line parallel to the y-axis through Q at the point R. Prove that R lies on the ellipse
    [tex]((x^2)/9) + ((y^2)/4) =1[/tex]

    Some of the facts I've established:

    the distance from P to Q is 1 for sure. Also, the triagle PQR is a right angle triangle. using the ellipse equation, i know that the distance from the center to the vertex is 3, and that the distance from the center to the focus is the square root of 5.
    Last edited: May 31, 2005
  2. jcsd
  3. May 31, 2005 #2
    Why wouldn't the distance of PQ be 5?
  4. May 31, 2005 #3
    if the radius of the inner circle is 2 and the outer one is 3, is the distance between them not 1?
  5. May 31, 2005 #4
    Yeah, for some reason I wasn't thinking about those being the squares of the radii. Anyway, I think [itex]R[/itex] would have the ascissa of [itex]Q[/itex] and the ordinate of [itex]P[/itex]. If you solve the respective equations for them, you find [itex](\sqrt{9-y^2}, \sqrt{4-x^2})[/itex]. I would try solve the equasion of ellipse for x and then y and it should be equivelent to the coordinates. I'm not very good at making proofs though, so there might be a problem with that way.
    Last edited: Jun 1, 2005
  6. Jun 1, 2005 #5


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    Write the point P as [tex](x_0,y_0)[/tex] and Q as [tex](x_1,y_1)[/tex]. The The fact that P and Q lie on the same radius means that [tex]\frac{y_0}{x_0}= \frac{y_1}{x_1}[/tex]. The horizontal line through P (parallel to the x-axis) is y= y0 and the vertical line through Q (parallel to the y-axis) is x= x1. The point R has coordinates (x1,y0). Of course, [tex]x_0^2+ y_0^2= 4[/tex] and [tex]x_1^2+ y_1^2= 9[/tex]. Thats a total of 3 equation is 4 unknowns. Use the equations to eliminate x1 and y0. I wouldn't be at all surprized if you were left with [tex]\frac{x_0^2}{9}+{y_1^2}{4}= 1[/tex]!
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