Finding Tangent Lines Through the Origin to Implicit Differentiation

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Discussion Overview

The discussion revolves around finding equations for two lines through the origin that are tangent to the implicit curve defined by the equation x^2 - 4x + y^2 + 3 = 0. Participants explore the application of implicit differentiation and the geometric interpretation of tangent lines in relation to the curve, which is identified as a circle.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in finding the tangent lines and mentions using implicit differentiation to find dy/dx.
  • Another participant questions the clarity of the original poster's understanding of points of tangency and the relationship of the origin to those points.
  • Some participants clarify that the tangent lines must intersect the curve at points other than the origin, as (0,0) does not satisfy the curve's equation.
  • A hint is provided that reformulates the curve's equation into a standard circle form, suggesting a geometric interpretation.
  • There is a discussion about the slope of the tangent line at a point of tangency and how it relates to the line through the origin.
  • A participant mentions using a guessed point to find the tangent line, indicating a trial-and-error approach.
  • Random tips are shared regarding implicit differentiation, emphasizing the importance of treating y as a function of x.
  • Some participants express confusion over the use of a constant 'r' and its relevance to the problem, leading to further clarification about the circle's radius.
  • One participant suggests solving a system of equations derived from the slope conditions and the curve's equation to find the points of tangency.

Areas of Agreement / Disagreement

Participants generally agree on the need to find tangent lines through the origin that intersect the circle, but there are multiple interpretations and methods proposed. The discussion remains unresolved as participants explore different approaches without reaching a consensus on a specific solution.

Contextual Notes

There are limitations in the understanding of the relationship between the origin and the points of tangency, as well as confusion regarding certain constants and their meanings. The discussion also highlights the need for clarity in the application of implicit differentiation.

Stalker23
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I can't get the problem. can anyone help me please.

-Find equations for two lines thorugh the origin that are tangent to the curve x^2 - 4x +y^2 + 3 = 0.

I found dy/dx=(-x+2)/y and put thta into the point slope equation, and then filled in (0,0) for the point, but couldn't get an equation.


any help appretiated.
 
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"I found dy/dx=(-x+2)/y and put thta into the point slope equation, and then filled in (0,0) for the point, but couldn't get an equation"

What point??
Please clarify to us, and not the least to yourself, what sort of points you're after, and what relation the origin has to those points!
 
Last edited:
I am looking for a tangent line that goes through the equation. the question i gave is all that is in the question. Only poing i think we know is that th eline goes through 0,0

exact question

Find equations for two lines thorugh the origin that are tangent to the curve x^2 - 4x +y^2 + 3 = 0.
 
Stalker23 said:
I am looking for a tangent line that goes through the equation. the question i gave is all that is in the question. Only poing i think we know is that th eline goes through 0,0

exact question

Find equations for two lines thorugh the origin that are tangent to the curve x^2 - 4x +y^2 + 3 = 0.

Hint: x^2 - 4x + y^2 + 3 = 0 => (x - 2)^2 + (y - 0)^2 = 1.
 
First, a tangent line does not go through an 'equation', it goes intersects a graph- you mean the tangent line goes through the graph of the relation x^2 - 4x +y^2 + 3 = 0 (that's a circle with center at (2, 0) and radius 1). You seem to be assuming that the lines must be tangent to the curve at the point (0,0) which can't be true: (0, 0) does not satisfy the equation. You are seeking two lines that pass through (0,0) and are tangent to the circle at some point on the circle.

Call the point of tangencey (x0,y0). Then any line through that point and (0, 0) has equation [itex]y= \frac{y_0}{x_0}x[/itex].
The slope is [itex]\frac{y_0}{x_0}[/itex]. For what (x0, y0) is the derivative of x2 - 4x +y2 + 3 = 0 equal to [itex]\frac{y_0}{x_0}[/itex]?
 
its a circle. how does putting the equation like that help me find the tangent line


my friend "guessed" a point where the circle looks like it starts to curve away from the origin, and then used that to solve for the tangent line, and he seems to have gotten the right answer, but we can't figure out how to get x= 1.5 allgebraically.
 
Random Tip: When doing implicit differentiation, it helps to replace "y" by "y(x)" so that it is clear that y is a function of x and not just another constant. Most beginners make the mistake of thinking of y as a constant and erroneously conclude that dy/dx = 0 (instead of using the chain rule).
 
Continue along the track shown you by HallsofIvy.
1. Assuming that [itex](x_{0},y_{0})[/itex] is a point of tangency on your curve, what is the slope at [itex](x_{0},y_{0})[/itex]?

2. Do you have a second expression for the very same slope, given the information that the tangent line at [itex](x_{0},y_{0})[/itex] goes through the origin?

3. Set the two expressions from 1. and 2. together as an equation!

4. What is the system of equations you need to solve for [itex](x_{0},y_{0})[/itex]?

5. Solve that system of equations!
 
Swapnil said:
Random Tip: When doing implicit differentiation, it helps to replace "y" by "y(x)" so that it is clear that y is a function of x and not just another constant. Most beginners make the mistake of thinking of y as a constant and erroneously conclude that dy/dx = 0 (instead of using the chain rule).

o yes. thankyou. our teacher made that point in class as well. I use a different symbol, but they have the same purpose.

i'm sorry but i still don't understand how to solve this equation. All of these hints are helping me get a little closer...but i still don't get it. Like, i have no clue where you guys are getting r from... (radius...?)

thanks for the help thus far.
 
  • #10
Perhaps you could start out with my list, then?
 
  • #11
Doing 1. and 2. for you:
1. You have already done this, sort of:
[tex]\frac{dy}{dx}\mid_{(x_{0},y_{0})=\frac{2-x_{0}}{y_{0}}[/tex]
2. HallsofIvy did this one for you: The slope of a line going through the origin AND [itex](x_{0},y_{0})[/itex] is:
[tex]\frac{y_{0}}{x_{0}}[/itex]<br /> <br /> Can you carry on now?[/tex]
 
  • #12
no, sorry, we've never used the constant r- what is it. is it the radius of the circle- correlation coeficiant??
 
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  • #13
Stalker23 said:
no, sorry, we've never used the constant r- what is it. is it the radius of the circle- correlation coeficiant??
Now you've confused me! I've looked back through all the responses and find no reference to a "constant r"! Did someone edit or delete it?

Any way, the point is that the circle given, [itex]x^2- 4x+ y^2+ 3= 0[/itex], which is equivalent to [itex](x- 2)^2+ y^2= 1[/itex] is a circle with center at (2, 0) and radius 1. It doesn't pass through (0,0) so you want to find two lines that do go through (0,0) and are tangent to the circle at two other points.

You could do this geometrically: The line segement from (0,0) to (2, 0), the line segment from (2, 0) to the point of tangency, and the line segment from (0,0) to the point of tangency form a right triangle with one leg of length 1 (the radius of the circle) and hypotenuse of length 2 (from (0,0) to (2, 0)) and so the length of the other leg is [itex]\sqrt{3}[/itex]. Any point (x,y) on the circle must satisfy [itex](x-2)^2+ y^2= 1[/itex] and if the line from (0,0) to (x,y) has length [itex]\sqrt{3}[/itex] then
also [itex]x^2+ y^2= 3[/itex]. Solve those two equations for the points of tangency.

But, using calculus, do what I suggested before. The line from (0,0) to (x, y) has slope [itex]\frac{y}{x}[/itex] and that must be the y' at the point on the circle. Using implicit differentiation, 2x- 4+ 2yy'= 0 so [itex]y'= \frac{4-2x}{2y}= \frac{2-x}{y}[/itex]. Solve [itex]\frac{2- x}{y}= \frac{y}{x}[/itex] along with [itex]x^2- 4x+ y^2+ 3= 0[/itex] for x and y. Those two methods should give the same point.
 
  • #14
Stalker23 said:
no, sorry, we've never used the constant r- what is it. is it the radius of the circle- correlation coeficiant??
Have you even bothered to read any replies you've been given?
Is independent thinking so difficult for you that you are unable to understand clear ideas given to you as hints?
 

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