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Find the four points at which two lines are tangent to both parabolas

  1. Jul 31, 2011 #1
    1. The problem statement, all variables and given/known data

    Draw a diagram to show that there are two lines tangent to both of the parabolas

    [itex]y = - x^{2}[/itex] (1)


    [itex]y = 4 + x^{2}[/itex] (2)

    Find the coordinates of the four points at which these tangents touch the parabolas.

    2. Relevant equations

    [itex]y - y_{o} = m(x - x_{o})[/itex]

    3. The attempt at a solution

    In creating a diagram, it appears that for such a scenario to occur one of the two lines must be tangent to one side (right will be used) of the parabola opening up (2) and then tangent to the other side (left), but of the parabola opening down (1) -- let this be line 1. Similarly, for the other line, it must be tangent to the other side (left) of (2), and then tangent to the other side (right) of (1) -- let this be line 2. So, it might appear that there are only two x-values for all four points (one negative, one positive) and two y-values (one negative, one positive), being that both curves are symmetrical at x = 0.

    In theory, I believe that for a line to be tangent to both parabolas at two points x = a [on (2)] and x = b [on (1)] (that is, considering only line 1), it must have only one slope. Therefore, letting (1) equal to f(x), and (2) equal to g(x),

    [itex]D_{x}g(a) = D_{x}f(b)[/itex]
    [itex]2a = -2b[/itex]
    [itex]a = -b[/itex]

    So, this scenario is true when the x-coordinates are symmetrical about x = 0.

    On line 1 tangent to (2) in the first quadrant at P(a, c) and tangent to (1) in the third quadrant at Q(b, d), the points can be expressed differently.

    [itex]P(a, c) = P(a, 4 + a^{2})[/itex]
    [itex]Q(b, d) = Q(-a, -[-a]^{2}) = Q(-a, -a^{2})[/itex] -- [itex]b = - a[/itex]

    On line 2 tangent to (2) in the second quadrant at U(e, f) and tangent to (1) in the fourth quadrant at S(g, h), the points can be express differently.

    [itex]U(e, f) = U(-a, 4 + [-a]^{2}) = U(-a, 4 + a^{2})[/itex]
    [itex]S(g, h) = S(a, -a^{2})[/itex]

    So, the slope of line PQ should be equal to either of the derivatives (the derivative of either curve) and the slope of line US should also be equal to either of the derivatives.

    For line one,

    [itex]\frac{4 + a^{2} + a^{2}}{a + a} = 2a[/itex]
    [itex]\frac{4 + 2a^{2}}{2a} = 2a[/itex]
    [itex]4a^{2} = 4 + 2a^{2}[/itex]
    [itex]2a^{2} = 2 + a^{2}[/itex]
    [itex]0 = 2 - a^{2}[/itex]
    [itex]a = \pm \sqrt{2}[/itex]

    It appears the plus or minus further confirms the symmetry.

    For line two,

    [itex]\frac{4 + a^{2} + a^{2}}{-a - a} = 2a[/itex]
    [itex]\frac{4 + 2a^{2}}{-2a} = 2a[/itex]
    [itex]-4a^{2} = 4 + 2a^{2}[/itex]
    [itex]-2a^{2} = 2 + a^{2}[/itex]
    [itex]0 = 2 + 3a^{2}[/itex]
    [itex]a = \pm \sqrt{-\frac{2}{3}}[/itex]

    Which is not defined (I found this odd).

    Therefore, the four points are,

    [itex]P(\sqrt{2}, 4 + [\sqrt{2}]^{2}) = P(\sqrt{2}, 6)[/itex] -- Because P is in the first quadrant, touching curve (2)
    [itex]Q(-\sqrt{2}, -[-\sqrt{2}]^{2}) = Q(-\sqrt{2}, -2)[/itex] -- Because Q is in the third quadrant, touching curve (1)
    [itex]U(-\sqrt{2}, 4 + [-\sqrt{2}]^{2}) = U(-\sqrt{2}, 6)[/itex] -- Because U is in the second quadrant, touching curve (2)
    [itex]S(\sqrt{2}, -[\sqrt{2}]^{2}) = S(\sqrt{2}, -2)[/itex] -- Because S is in the fourth quadrant, touching curve (1)

    And, the points illustrate symmetry about x = 0 (spefically, P and U, and, Q and S).


    Much appreciation for any help!
    Last edited: Jul 31, 2011
  2. jcsd
  3. Aug 1, 2011 #2
    I used different unknowns, but exactly the same results. Now all you really have to do is plug them back into f(x) and g(x) to get the y coordinates of the points and you're done.

    Could you explain why you have the equation equal to 2a? Line 2 is basically going to be the same as line 1 except that their slopes will differ by a negative. I worked it out using [tex]\frac{y_2-y_1}{x_2-x_1}[/tex] and knowing that it's equal to the derivatives of the functions at x1 and x2. When I went through this the second time for line 2, I have everything set up exactly as you, but the right side of the equation turned out to be -2a. This will give you ±√2 as in line 1, as it should.

    It's kinda hard to keep track of all the letters without a graph for reference (plus I need to be going to bed..., and there might be something I missed) so I couldn't justify all your steps for working out line 2, but not having -2a is what's giving you that ±√(-2/3). See if you can work that out again. But you did get the correct points in the end!
  4. Aug 1, 2011 #3


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    It should be obvious from the symmetry that the two lines must cross at (0, 2), the point midway between the two vertices. That, in turn, means the lines must be of the form y= mx+ 2.

    Now, it y= mx+ 2 is tangent to [itex]y= x^2+ 4[/itex], then we must have [itex]x^2+ 4= mx+ 2[/itex] at that point, of course, and, to be tangent, that x must be a double root. Use the quadratic formula to determine what m must be in order that this equation have a double root.
  5. Aug 1, 2011 #4
    To start, thanks for your help!

    My reasoning for setting the equation for line 2 equal to 2a is that I know that both lines touch both curves, meaning that its slope should be equal to the derivatives of both curves. So, I should be able to use either derivative as the slope for the right side of the equation. And, as I *just* now realized, being that this is line 2, it has points U(-a, 4 + a2) and S(a, -a2), therefore resulting in a negative slope -- I had accidentally considered line 2 as line 1 in those or more calculations. Hahaha, well, there you go! (And I even had a graph to look at! ;P)

    I am not sure if I understand: is everything generally okay, or is my solution a bit wonky (I would like it to be proper)?
    Last edited: Aug 1, 2011
  6. Aug 1, 2011 #5
    To begin, thanks for your help!

    If its not much of a bother, could you explain the reasoning behind the statement of how the two lines must cross at (0, 2), the point midway between the two vertices? I was thinking that the two lines had an equal y-intercept, but I could not provide a proof of why, let alone explain why. However, I do understand the general linear equation you concluded with.

    Ah, interesting! I will certainly work through this method to have another method of solution to this problem.
  7. Aug 1, 2011 #6
    I was basically saying that you worked it out great, but that not having -2a was giving you ±√(-2/3).

    Well since the two parabolas have the same shape, the lines that are tangent to them will have some symmetry as a result. Take a look at a graph of the two parabolas, and the tangent lines too if you can. The tangent lines will go through the point right in between the vertices, which is (0, 2).
    This suggests that (0, 2) is the y-intercepts for both the tangent lines. If you plug everything into the equation for a tangent line at some point a and simplify it to the form y = mx + b, you will get b = 2. Once I got a = ±√2, I plugged them into y = f'(a)(x - a) + f(a) and got 2 for the y-intercept.
  8. Aug 3, 2011 #7
    Oh, okay! And, very interesting! I do think I understand now. Thanks for all your help!
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