How Do You Find a Common Tangent Line to Two Parabolas?

Click For Summary

Discussion Overview

The discussion revolves around finding a common tangent line to two parabolas defined by the functions \( f(x) = x^2 - 3 \) and \( g(x) = (x - 3)^2 \). Participants explore the conditions under which a line \( T \) can be tangent to both parabolas, addressing various parts of the problem including the relationship between the points of tangency, the slope of the tangent line, and the setup for calculating the area between the curves.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant suggests that the slope of the tangent line must equal the derivative of the parabolas at the points of tangency, indicating a relationship between \( a \) and \( b \) as \( a = b - 3 \).
  • Another participant proposes using the form \( y = mx + b \) for the tangent line and setting the discriminants of the resulting equations to zero to find \( m \) and \( b \).
  • A different participant expresses uncertainty about the formulation of the tangent line, suggesting it may take the form \( 2x(x - a) + (a^2 - 3) \) and \( 2x(x - b) + (b - 3)^2 \).
  • One participant provides a detailed approach to derive the equations for the tangent line, leading to the conclusion that \( m = 1 \) and \( b = -\frac{13}{4} \), thus proposing the tangent line equation \( y = x - \frac{13}{4} \).
  • Another participant acknowledges the help received and expresses intent to continue working on the problem.

Areas of Agreement / Disagreement

Participants present various methods and approaches to find the tangent line, but there is no consensus on the final solution or the derivation process. Some participants express uncertainty about specific steps, indicating that the discussion remains unresolved.

Contextual Notes

Participants rely on the discriminant condition for tangency, but the discussion does not fully resolve the implications of the derived equations or the numerical values of \( a \) and \( b \). There are also assumptions about the forms of the equations that may not be universally accepted.

karush
Gold Member
MHB
Messages
3,240
Reaction score
5
https://www.physicsforums.com/attachments/2227
The region $R$, is bounded by the graphs of
$$ f(x)=x^2 -3$$ and $$g(x)=(x-3)^2$$,
and the line $$T$$, as is shown in the figure above.
$$T$$ is tangent to the graph of f at point $$(a,a^2-3) $$
and tangent to the graph of g at point $$(b,(b-3)^2)$$

a. Show that $a=b-3$
b. Show the numerical value of $a$ and $b$
c. Write the equation of $T$
d. Setup but do not calculate the integral for the region $R$.

this is a common problem, and I looked at several solutions via several forums but only got lost on this. I know that the slope of the $2$ parabola's is $2x$ and since it is a line the slope is the same. I did try using $y=mx+b$ but with $3$ variables $a,b$ and $x$ it was hard work with. If I could see how the equation of $T$ is derived then the other questions I should be able to answer.
the answer is not given but the line eq appears to be close to $y=x-3.25$:cool:
 
Physics news on Phys.org
Re: tangent line to 2 parabola’s

What I would for part c) is let the tangent line be $y=mx+b$. Next, equate this line to the two parabolas in turn and require the resulting discriminants to both be zero, giving two equations in two unknowns from which $m$ and $b$ can be easily found. Once you have this, then the other parts of the question should be a piece of cake. :D
 
Re: tangent line to 2 parabola’s

not sure if I follow ...
but presume $T$ is both $2x(x-a)+(a^2-3)$ and $2x(x-b)+(b-3)^2$
 
Re: tangent line to 2 parabola’s

What I did was:

i) $$x^2-3=mx+b$$

standard form...

$$x^2-mx-(b+3)=0$$

discriminant to zero...

$$m^2+4b+12=0$$

ii) $$(x-3)^2=mx+b$$

standard form...

$$x^2-(m+6)x+(9-b)=0$$

discriminant to zero...

$$m^2+12m+4b=0$$

Now, since both discriminants are zero, we may equate them and write:

$$m^2+4b+12=m^2+12m+4b$$

Collect like terms:

$$12m=12$$

Divide through by 12:

$$m=1$$

And thus, we can find from either equation:

$$b=-\frac{13}{4}$$

Hence, the tangent line is:

$$y=x-\frac{13}{4}$$
 
Re: tangent line to 2 parabola’s

OK, I see now, will try to get the rest of it. that was a big help
BTW I have an incredible view of Pearl Harbor here at LCC:cool:
 
Last edited:

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K