Points on ellipse where tangent slope=1

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Homework Help Overview

The problem involves finding points on the ellipse defined by the equation x² + 2y² = 1 where the slope of the tangent line equals 1. Participants are exploring the relationship between the coordinates and the conditions for the tangent slope.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the correct pairing of x and y values derived from the ellipse equation and the condition for the tangent slope. There is uncertainty about how to determine which coordinates correspond to valid points on the ellipse.

Discussion Status

Some participants have provided insights into the relationship between the slope condition and the resulting coordinates, noting that only two points satisfy the tangent slope condition despite initially finding four possibilities. The discussion includes suggestions to visualize the problem through diagrams to clarify the situation.

Contextual Notes

Participants mention the importance of accurately representing the problem through diagrams and express concerns about potential test implications. There is an acknowledgment of the challenge in reconciling the algebraic results with the geometric interpretation.

Feodalherren
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Homework Statement



7
Find the points on the ellipse x^2 + 2y^2 = 1 where the tangent line has slope 1

Homework Equations





The Attempt at a Solution


I got the correct X and Y values but this gives me four possibilities and the answer key says there are two points.

I got x = +/- sqrt 2/3 and y = +/- sqrt 1/6
The values are correct but how am I supposed to know which coordinate to pare with which?

For an example, how do I know if it's (-sqrt 2/3, -sqrt 1/6) or (sqrt 2/3, -sqrt 1/6) ?
 
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Feodalherren said:
I got x = +/- sqrt 2/3 and y = +/- sqrt 1/6
The values are correct but how am I supposed to know which coordinate to pare with which?

For an example, how do I know if it's (-sqrt 2/3, -sqrt 1/6) or (sqrt 2/3, -sqrt 1/6) ?

It helps to draw a diagram. If the slope is +1, aren't there only two possible tangents? In which quadrants do their points of intersection with the ellipse lie?
 
From x^2+ 2y^2= 1 you get 2x+ 2yy'= 0 so that y'=- x/y= 1. That gives y= -x and putting that into the equation of the ellipse, x^2+ 2x^2= 3x^2= 1 so that x= \pm\sqrt{3}/3.

That's what you got , right? And that gives four points on the ellipse?

But your condition is y= -x. Putting x= \sqrt{3}/3 and x= -\sqrt{3}/3 into that gives only two points.

(The other two points on the ellipse give slope -1.)
 
Oh, DUH! Thanks guys :D.
 
Sunil Simha said:
It helps to draw a diagram.
It's hard to overemphasize the importance of this advice. A halfway decent diagram can give you insight that you can't get just fiddling with equations.
Sunil Simha said:
If the slope is +1, aren't there only two possible tangents? In which quadrants do their points of intersection with the ellipse lie?
 
I actually did draw a diagram the first time I did it but I still got four points, it just didn't strike me to draw the tangent lines. If this question is on my test today (in 1.5hrs) I'll nail it. Thanks again.
 
Feodalherren said:
I actually did draw a diagram the first time I did it but I still got four points, it just didn't strike me to draw the tangent lines.
Try to draw a graph that represents the situation; namely, an ellipse with some points where the slope of the tangent line equals 1. If you got four points, your drawing wasn't an accurate representation of the question.
 
I realize that now. It became painfully obvious as soon as I read Mr Simha's answer.
 

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