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Y=x*y=x yields y=x^2 tangents

  1. Feb 15, 2006 #1
    Ok, I need to find 2 lines that when multiplied (y=x*y=x yields y=x^2)
    create a parabola in which those two lines are tangent to it. My problem is that I have no idea where to start. I think that the two lines have to have opposite slopes, or at least one slope must be positive and one must be negative, else both lines would follow the same direction, and there is an obvious problem there...
  2. jcsd
  3. Feb 15, 2006 #2


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    This was surprisingly tricky. The first thing to realize is that if a parabola can be expressed as the product of two lines ax+b and cx+d, then the parabola's equation is (ax+b)(cx+d) and the lines factor it. The next thing to realize is that the parabola must touch a line at each zero of the parabola, and if the line is not tangent at the zero it is not tangent anywhere. Now differentiate (ax+b)(cx+d) (it's easier later on if you do it using the product rule rather than expanding it out first). Then, you must find a simple condition to make the slope of the parabola at the zeroes appropriate, and then solve that condition.

    There is also the trivial case of y1 = 0 = y2 but since you ask for a parabola I don't think that is a valid answer.

    Edit: it's interesting, the two lines must always intersect at the same y coordinate, 0.5.

    What class is this for?
    Last edited: Feb 15, 2006
  4. Feb 15, 2006 #3
    It's for my Honors Algebra II class. Not everyone has to do it, my teacher and I are working together on upper level math like this. I can generally get the stuff we discuss, but some of it goes right over my head, as I'm only in 9th grade. But thanks alot for the help on this one, it's greatly appriceated.
  5. Feb 15, 2006 #4


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    Well, if you're only in algebra II I don't see how you can do it without differentiating. Are you given a formula to tell if a line is tangent to a parabola?
  6. Feb 15, 2006 #5
    Nope, and that's where I'm stuck at the moment. I see the (ax+b)(cx+d)=y, but I've got nothing after that... I tried expanding it out which gives
    y=acx^2+adx+bcx+bd and that yields nothing that I can see.
  7. Feb 15, 2006 #6


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    If you don't know how to take a derivative and you have not been given a formula to tell when a line is tangent to a parabola, you can't do the problem.
  8. Feb 15, 2006 #7
    Ok, thanks anyway then. I guess I'll look for that formula, as I took a look at derivatives, and I doubt I could learn that without having a foundation in Calculus let alone Trig or Euclidian Geometry.
  9. Feb 15, 2006 #8


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    You don't need calculus to find out when a line is tangent to a parabola. You just need to know how many times they intersect.
  10. Feb 15, 2006 #9


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    Oh, right! It is also much simpler to do it from that perspective.
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