1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor

    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]!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook