Common tangent to two parabolas

[apple2357]

I am trying to explore whether it is possible to find a common tangent to two parabolas? And i can't work out if it is always possible.

I started with these two:

y= (x-2)^2 +3
and
y= -(x+2)^2 -3

By looking for a value at which the gradient of a tangent will be the same, and some messing with algebra, I established the line y= (sqrt28- 4)x is a common tangent to both parabolas.

Now, i can't work out if it is only possible to find common tangents to parabolas only if one 'upside down', like my example above?
My algebraic method fails if they are not, but visually I feel it must be possible?

( By common tangent i mean if one parabola has a tangent line at a particular point , the same line is also a tangent to the other parabola at some point on the curve..hope that makes sense)

Any thoughts?

 

[apple2357]

Ok, i think i have solved this. So it is always possible to find a common tangent.
Since posting i have refined by first method.
If you can offer an interesting method other than algebraic ( i used the discriminant and solved some equations by assuming the line is y=mx+c) please suggest below!

 

[Dr.Peterson]

Ok, i think i have solved this. So it is always possible to find a common tangent.
Since posting i have refined by first method.
If you can offer an interesting method other than algebraic ( i used the discriminant and solved some equations by assuming the line is y=mx+c) please suggest below!

Certainly it isn't always possible; consider y=x^2 and y=x^2+1, where one lies entirely within the other; or, when they open in opposite directions, I think it would be impossible if they intersect, like y=x^2 and y=1-x^2. But it would usually be possible, and you could find specific conditions. There will also be cases when there are two common tangents.

I haven't given any time to trying to actually solve it, only imagined the graphs. The set of all tangent lines to a given parabola passes through every point outside the parabola, and we can picture sweeping the tangent across the other parabola until it becomes tangent.

Would you like to show your general results?

 

[Dr.Peterson]

I did the algebra, keeping it simple by assuming one parabola is y=x^2, and allowing the other to be y=a(x-h)^2+k. (Presumably you, like, are assuming the axes of both parabolas are vertical; much more happens in the general case.) This special case covers all cases if you do some transformations.

I found the discriminant for solving for the intersection of a tangent to the first parabola, with the second parabola, which yielded a quadratic in m (from which we would find the slope of the common tangent(s)); the discriminant of that turned out to be identical to the discriminant that determines how many intersections the two parabolas have. The result is exactly what I predicted: if a>0, there is no tangent when the parabolas don't intersect; and if a<0, there is no tangent when they do intersect.

There's more to be explored -- how to tell when this happens based on a, h, k; when there are two common tangents; and maybe more.

I didn't try a more calculus-based approach, so I can't say whether that is any easier, but this wasn't too bad.

 

[apple2357]

Thanks very much. I rushed in a bit too much when i said 'always works' !
It's quite a pretty problem but i couldn't find a calculus approach either.

 

[JeffM]

I have been busy on other things and not had time to work out proofs. But it is obvious that every parabola that has an extremum with a value of n is tangent to the line h(x) = n. So, a common extremum entails at least one common tangent. The next step would be to show that a common extremum precludes any other common tangent (assuming that is true of course).

What I have not yet found a proof for is that if f(x) is a parabola, z is positive, and f(x) = z - g(x), there are exactly two distinct lines that are tangent to both f(x) and g(x). I currently believe that to be true, but, because I have not had time to create a proof, I may be wrong.

I suspect that unless two parabolas share an extremum or an axis, they do not share a tangent, but I am very far from being able to do more than wave my hands on that point. As I have time, I shall try to give updates. I am no mathematician and am slow developing proofs.

 

[JeffM]

\(\displaystyle \text {Given: } f(x) = ax^2 + bx + c,\ a \ne 0,\ g(x) = ix^2 + jx + k,\ i \ne 0, \text { and } f(x) \not \equiv g(x).\)

\(\displaystyle \text {Theorem 1: } \dfrac{4ac - b^2}{4a} = n = \dfrac{4ik - j^2}{4i} \implies\)

\(\displaystyle h(x) = n \text { is a tangent to both } f(x) \text { and } g(x).\)

\(\displaystyle \text {Proof: } h(x) = n \text { and } h'(x) = 0 \text { for all real values of x.}\)

\(\displaystyle f'(x) = 0 \iff x = -\ \dfrac{b}{2a}.\)

\(\displaystyle f \left (-\ \dfrac{b}{2a} \right ) = a * \left ( - \dfrac{b}{2a} \right )^2 + b * \left ( -\ \dfrac{b}{2a} \right ) + c =\)

\(\displaystyle \dfrac{ab^2}{4a^2} - \dfrac{b^2}{2a} * \dfrac{2}{2} + c * \dfrac{4a}{4a} = \dfrac{4ac - b^2}{4a} = n = h(x).\)

\(\displaystyle \therefore h(x) \text { is a tangent to } f(x) \ \because \ f(x) = h(x) \text { and } f'(x) = h'(x) \text { at } x = -\ \dfrac{b}{2a}.\)

\(\displaystyle \text {Similarly, } h(x) \text { is a tangent to } g(x) \text { at } x = - \ \dfrac{j}{2i}.\)

\(\displaystyle \text {THUS, } h(x) \text { is a tangent to both } f(x) \text { and } g(x) \text {. Q.E.D.}\)

This just proves what was obvious to anyone who knows a little calculus: namely that all parabolas with the same extremum have at least one tangent in common. I would not want to try this proof without calculus.