How Do You Calculate the Intersection of a Line and an Ellipse?

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Homework Help Overview

The discussion revolves around calculating the intersection of a line and an ellipse defined by the equation b²x² + a²y² = a²b², where the ellipse is centered at the origin. The original poster expresses confusion about deriving the coefficients A, B, and C for the quadratic equation after substituting the line's equation into the ellipse's equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss expanding the squared term from the substitution and combining like terms to achieve the quadratic form. There are questions about the correctness of the distribution and the resulting coefficients.

Discussion Status

Some participants have offered guidance on expanding the equation and simplifying it, while others have raised concerns about potential mistakes in the original calculations. The conversation reflects a mix of attempts to clarify the process and explore the implications of the intersection conditions.

Contextual Notes

There is an assumption that the problem involves an arbitrary line, which may lead to different interpretations of the intersection scenario. Participants also mention the importance of understanding tangential conditions and the behavior of the line relative to the ellipse.

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Compute the intersection of a line and an ellipse centered at (0,0). Ellipse equation is b²x² + a²y² = a²b²
where b is the minor axis and a is the major axis.

I am having trouble finding A, B, and C- somewhere down the line, I know the Quadratic Eq. is used to find x. I also know AX² + BX + C = 0 is used. I've combined the slope (y = mx + c) into the ellipse equation ==> b²x² + a² (mx+c)² = a² b². I then distributed the a², yielding b²x² + (a²mx + a²c)² = a²b². Here is where I'm stuck. I know that ultimately I am trying to find A, B, and C.

A= (a²mx + a²c)²??
B= ??
C= b²x² - a²b²?? I am SO lost and confused! lol
 
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You can expand the (a2mx+a2c)2 and then combine like (x) terms. This will get you the quadratic form you are looking for
 
OK, so now I have 2a^4mx² + 2a^4mxc + 2a^4c². This doesn't seem to allow me to combine anything.
 
1. I would preface my answer with "assuming this means the intersection of an arbitrary line with this ellipse". Because it seems to me easy to read the question as asking about the line through the origin - an easier problem.

2. I am sure I make more mistakes than Office Shredder but I think he has made one there and you have carried it forward.

3. However I do see no alternative to expanding that square term as he says and slogging out the resulting quadratic. Try to simplify or take out factors as much as possible at every stage. It is just a matter of being careful but a routine with nothing that should be new to you.

4. After that I advise take a leaf from the book of Polya 'How to Solve It.' He says, when you have solved your problem it's not finished. Try and answer whilst you are warm what is the condition of the line to be tangent to the ellipse? When does it cut the ellipse at all and when not? What are the co-ordinates of the tangent point? The equation of the tangent line? its slope? Keep notes maybe. Because these questions will sure come up in future lessons or exercises maybe soon, and you will have put yourself in a better and more confident position when it does, apart from starting a useful habit for always.
 
Last edited:
epenguin said:
2. I am sure I make more mistakes than Office Shredder but I think he has made one there and you have carried it forward.

In my defense I just copy/pasted from his post! :-p

But yeah, looking closer you didn't distribute the a2 correctly. Once you've expanded the quadratic, you need to plug it back into your equation and solve for x
 

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