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# Can you solve this?

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Ok, a student from a Calculus I class asked if I knew how to solve this problem: Find two shared tangents lines of the graphs y = x^2 and y = -x^2 + 2x - 5. This question seemed easy at first, but I am having trouble solving it. I''ve never had to solve something like this before. He was just looking at the difficult problems at the end of the chapters for fun, and he didn''t know how to go about solving this question. Anybody have any idea how you would find shared tangent lines on two parabolas?

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I''d make an expression for the tangent line based on where it tangents the first parabola(at x = a), then solve a by making sure the gradient of the tangent in the intersection between the other parabola and the line are equal. There should also only be one intersection between them.

But that''s just me. I haven''t taken calculus yet. I''ll see what university yeilds.

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let me try:

...make them equal each other (wrt to y)..

...we get

2x^2 - 2x + 5 = 0;

...then do the equation: x = (-b + sqrt(b^2 - 4ac)) / (2a)

x = (2 + sqrt(4 - 20)) / 4  = (2 + sqrt(-16)) / 4  = (2 + 4i) / 4  = 1/2 + i

...hmm ok I am wrong somewhere..

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This should get you started:

Line equation is y = mx + b;
Take the derivative of the first parabola, call it f1(x). At the tangent point with the line (y1, x1), m = f1(x1).
Take the derivative of the second parabola, call it f2(x). At the tangent point with the line (y2, x2), m = f2(x2).
And of course (y1, x1) and (y2, x2) are related by the line equation.

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If by shared tangents you mean, that the lines ''cross'' and at that point are tangent that means that the first derivatives are equal (tangent), and the functions themselves are equal (shared point).

y1 = x*x
y2 = -x*x + 2x - 5

y1'' = 2x
y2'' = -2x + 2

Calculating the tangents first:

2x = -2x + 2
4x = 2
x = 1/2 --> y'' = 1

There is only one place where the graphs are tangent at the same x value: at x = 1/2 --> y1 = 0.25, y2 = -4.25. (Note the y values aren''t the same here)

If you mean any two tangent lines, that''s more difficult. The process would go the same if the lines may be anywhere, but then you''d have an x1, x2, y1 and y2, and you''d probably end up with mutliple solutions.

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If I am not mistaken you just need find the derivatives of the two functions and then find the places where they are the same.

So
y'' = 2x and
y'' = -2x + 2

Which would yield
2x = -2x + 2
4x = 2
x = 0.5

And so the lines would have the same tangent at x = 0.5.

Am I mistaken?

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Shared tangent means a line that is tangent for both curves (but it may have two different points of intersections).
The solution is straight forward and only takes a few lines.

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quote:

If by shared tangents you mean, that the lines 'cross' and at that point are tangent that means that the first derivatives are equal (tangent), and the functions themselves are equal (shared point).

I can't for the life of me figure out what this is supposed to mean. Could you clarify?

I'd personally interpret "shared tangent" as a line that is a tangent to both x^2 and -x^2 + 2x - 5, but not necessarily at the same point. Operating under this assumption, I solved the problem and produced the following picture: http://w1.315.telia.com/~u31523890/screen.jpg

Write down the equation for a tangent to y = x^2 at (a, a^2). Do the same for a tangent to y = -x^2 + 2x - 5 at (b, -b^2 + 2b - 5). What can you say about those equations, if you want a shared tangent?

[edited by - Muzzafarath on March 4, 2004 12:59:09 PM]

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Nice plot! And you got the same solution as I did!

Great

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