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## #1 2006-10-25 06:38:38

Mac "The Man" Smith-XNQBL
Guest

### Common Tangent Lines

Hi, im having problems with the following problem. Any help would be great, thanks!

Find the two points on the curve y = x^4 - 2x^2 - x that have a common tangent line.

## #2 2006-10-26 00:37:11

Toast
Real Member
Registered: 2006-10-08
Posts: 1,321

### Re: Common Tangent Lines

Find the derivative:

The curve of the derivative has an infinite number of coordinates which share the same 'y' values.

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## #3 2006-10-26 01:10:08

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: Common Tangent Lines

True, but that wouldn't necessarily mean that they'd have a common tangent. For that, the tangents would need to have the same y-intercept as well as the same gradient.

This is quite a tricky and interesting problem, actually. I'll come back to this in a while.

Why did the vector cross the road?
It wanted to be normal.

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## #4 2006-10-26 02:14:07

gnitsuk
Member
Registered: 2006-02-09
Posts: 121

### Re: Common Tangent Lines

therfore

So equation of a tangent to the curve at any point x is:

replacing y from original equation gives:

which gives intercept as

So say we have two points which have a common tangent, (a,b) and (c,d) then the x coordinates will have to simultaneously satisfy:

and

So we "just" have to solve these for a and c where a is not equal to b.

Ok, then I "cheat" and use mathematica:

eqns = {4a^3 - 4a - 1 == 4c^3 - 4c - 1, -3a^4 + 2a^2 == -3c^4 - 2c^2}
Solve[eqns, {a, c}]

Whicih gives six solutions including:

and

The others are similarly unpleasant.

This solution gives a is approx. 0.853 and c is approx 0.247 and if you look at the graph of the function at these points:

Plot of x^4-2x^2-x

This looks reasonable.

Last edited by gnitsuk (2006-10-26 02:22:37)

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## #5 2006-10-26 03:24:20

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: Common Tangent Lines

Wow. That was tricky. From the bit where you needed to solve those simultaneous equations, you could do it by writing the second equation in the form (pa²+q)²+r and then writing that in terms of a, then substituting into the first equation and solving for c.

But that would involve a ton of work and Mathematica is a far easier option.

Why did the vector cross the road?
It wanted to be normal.

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