Find the equation of the line tangent to two parabolas

  • Thread starter Thread starter flemonster
  • Start date Start date
  • Tags Tags
    Line Tangent
Click For Summary
SUMMARY

The discussion focuses on finding the equation of a line tangent to the parabolas defined by f(x) = x^2 - 2x + 2 and g(x) = -x^2 - 2x - 2. The user attempts to equate the slopes of the tangent lines derived from the derivatives of the parabolas, resulting in the equation 2x_1 - 2 = -2x_2 - 2. The user struggles with simplifying the equations to express them in terms of a single variable but ultimately finds a helpful tutorial from the University of Washington that aids in resolving the problem.

PREREQUISITES
  • Understanding of calculus concepts, specifically derivatives and tangent lines
  • Familiarity with quadratic functions and their properties
  • Ability to manipulate algebraic equations and solve for variables
  • Knowledge of the relationship between points on a curve and their respective equations
NEXT STEPS
  • Study the concept of tangent lines to curves in calculus
  • Learn how to derive equations from given functions and their derivatives
  • Explore the method of substitution in solving systems of equations
  • Review resources on quadratic functions and their graphical interpretations
USEFUL FOR

Students studying calculus, particularly those focusing on tangent lines and quadratic functions, as well as educators looking for examples of problem-solving strategies in mathematics.

flemonster
Messages
11
Reaction score
0

Homework Statement


Given the two parabolas: f(x) = x^2 - 2x + 2 and g(x) = -x^2 - 2x - 2. Find the equation of a line that is tangent to both curves.

Homework Equations



The given parabolas, equation for a line y = mx + b, and the derivatives of the two parabolas 2x - 2 and -2x - 2

The Attempt at a Solution



The line tangent to the two parabolas will pass through the points

(x_1 , y_1)

for the parabola f(x) and

(x_2 , y_2)

for the parabola g(x)

so the equations for the two lines will be,

for f' y_1 = (2x_1 - 2)x_1 +b

and

g' y_2 = (-2x_2 - 2)x_2 + b.

Since the slopes of both lines will be the same I thought that setting the two slopes equal might get me started so I wrote

2x_1 - 2 = -2x_2 - 2

which gave me \frac{x_1}{x_2} = -1.

I rearranged the two linear equations and set them equal:

y_1 - (2x_1 - 2)x_1 = y_2 - (-2x_2 - 2)x_2

but that got me absolutely nowhere. I got the whole thing down to

x^2 _1 + x^2 _2 = \frac{y_1 - y_2}{2}

but that doesn't help.

I know I need to limit my variables and try to get the whole thing in terms of one variable but I'm at a loss as to how to make that happen. Every time I substitute and simplify I get either one or negative one which tells me nothing. I can't figure out how relate the equations and simplify. Any help is appreciated.
 
Physics news on Phys.org
http://www.math.washington.edu/~aloveles/Math124Fall2012/ConroytangentLines01.pdf

I don't know how to do these problems, but that looks exactly how I would go about it. Sorry I can't give any additional help.
 
You do not appear to have used the fact that the points (x1, y1), (x2, y2) lie on the respective parabolas.
 
haruspex said:
You do not appear to have used the fact that the points (x1, y1), (x2, y2) lie on the respective parabolas.

I'm not sure what you mean. I accounted for them being on different parabolas with the subscripts but is there something in their being on different parabolas that I missed?
 
And I just figured it out. That tutorial from UW did the trick which is kind of funny seeing that's where I did my undergrad. Thanks iRaid!
 
flemonster said:
I'm not sure what you mean.
I mean that e.g. x1, y1 satisfy the equation of the first parabola.
 

Similar threads

Replies
15
Views
3K
Replies
2
Views
1K
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K