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Homework Help: from a question on conics

  1. Jun 4, 2006 #1
    I'm doing a question in my book on conics where there is a circle cutting through a parabola. There are three points S - the focus P - a point on the parabola and Q - the point where the tangent at P meets the directrix.

    The focus is at S[1,0] and the dirextrix is x=-1
    point P is [t^2, 2t] and point Q is [-1, (t^2 -1)/t]

    The question asks you to find the circle on which all three points lie.

    I said that the equation of the circle would be (x-a)^2 + (y-b)^2=c^2

    I put in the values of X and Y for each of the coordinates to get three expressions which are all equal to c^2 and then equated coefficients to get the values of a, b, and c. I thought this seemed obvious but my value for a keeps coming out as 1, which is wrong.

    Can anyone see any other solution?
     
  2. jcsd
  3. Jun 4, 2006 #2

    HallsofIvy

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    You say that " a point on the parabola" but then say that "P is [t^2, 2t]" which could be any point on the parabola. I think you mean that P and Q are points such that the tangent to the parabola intersects the directrix at Q.

    The parabola with focus at (1, 0) and directrix x= -1 has equation x= (1/4)y2. Differentiating with respect to x, 1= (1/2)y y' so
    y'= 2/y. If the point P is given by (t2,2t) then y'= 2/2t= 1/t.
    The equation of the tangent line at (t2,2t) is y= (1/t)(x- t2+ 2t. That intersects the directrix, x= -1 when y= (1/t)(-1-t2)+ 2t= -1/t- t+ 2t= t- 1/t = (t2-1)/t. That is, the values of P and Q given could be for any point on the parabola! You could evaluate the equation of the circle through S, P, and Q as a function of t.

    Write the equation of the circle as (x-a)2+ (y- b)2= c2 and substitute the coordinates of the three points for x, y:
    [tex]S: (1- a)^2+ b^2= c^2[/tex]
    [tex]P: (t^2- a)^2+ (4t^2- b)^2= c^2[/tex]
    [tex]Q: (1- a)^2+ (\frac{(t^2-1)^2}{t^2}-b)^2= c^2[/tex]

    Subtracting the equation for S from the equation for t eliminates both a and c and gives
    [tex]\frac{(t^2-1)^2}{t^2}-\frac{2(t^2-1)^2}{t^2}b= 0[/tex]
    so b= 1/2. (independent of t!).
    Putting that into the equation for S, c2= 5/4-2a+ a2. Now put that into the equation for P to determine a. Looks to me like a will be 4th degree polynomial in t.

    Another way to do this is to use this geometric property: all perpendicular bisectors of chords of a circle intersect at the center of the circle.
    The the points S, P, Q give two chords:
    The line segement from S to P: has midpoint
    [tex](\frac{t^2+1}{2},\frac{t}{2})[/itex]
    and slope
    [tex]\frac{t}{t^2-1}[/tex]
    and, so, the equation of the perpendicular bisector is
    [tex]y= \frac{1-t^2}{t}(x- 1)[/itex]

    The line segment SQ has midpoint
    [tex](0, \frac{t^1-1}{2t})[/tex]
    and slope
    [tex]\frac{t^2-1}{2t}[/tex]
    and, so, the perpendicular bisector has equation
    [tex]\frac{2t}{1-t^2}(x-1)[/itex]
    Solving those two equations simultaneously for x and y will give the coordinates a and b in terms of t. Of course, the distance from that point to (1, 0) will be c.
     
  4. Jun 4, 2006 #3
    ahhh, i had completely forgotten about the perpendicular bisector approach, thanks for reminding me. I'll try it now.

    Although I'm still not sure about the equation of a circle approach. I disagree with some of your working for that way. In subsituting values in you've written your Q x coordinate as 1 instead of -1, and you've squared all your y coordinates before putting them into the brackets. Surely both b and a would be variables?
     
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