Find Dimensions & Location of Box for Tilted Ellipse x^2 -xy +y^2 =3

[physstudent1]

Homework Statement

The graph of the tilted ellipse x^2 -xy +y^2 =3 is shown to the right. What are the dimensions and the location of the box containing the ellipse?

Note the sides of the box are vertical and horizontal and also are tangent to the elipse.

(The image is simply a tilted elipse inside a box which looks to be a square and is tangent to the elipse at four places two at the top right and two at the bottom left.

Homework Equations

The Attempt at a Solution

My first thought is to find where the derivative equals 0. The derivative I found through implicit differentiation is (2x-y)/(x-2y) = dy/dx I set this to zero but got an equation with two variables? I'm not sure what to do from here.

 

[D H]

In addition to the equation 2x-y=0 (the denominator is irrelevant) obtained by setting dy/dx to zero, you do have another equation at hand: the equation of the ellipse.

You have four lines to find: the four sides of the box. You will find two of these lines by setting dy/dx to zero and solving for the appropriate variable. You need to do something else to find the other two lines.

 

[atqamar]

When you set the impartial derivative equal to 0, you will only find the upper and lower bounds of the ellipse. After this step, you need to set the derivative equal to a different value to find the right and left bounds. How would you do that?

However, you impartial derivtive \frac{dy}{dx}=\frac{2x-y}{x-2y}=0 can be simplified... Simply solve for one variable, like y. Then, plug this value into your original equation x^2 -xy +y^2 =3 to find the value for one variable. Its basically a system of equations, with your original equation as part of the system.

 

[physstudent1]

I found 2x=y from the derivative plugged this back into the equation of the original ellipse to and found x=+1 and x=-1 then I used x=y/2 from the derivative plugged this back into the ellipse equation and found y = +2 and y = -2

 

[atqamar]

Right, so your coordinates for \frac{dy}{dx}=0 would be (-1,-2) and (1,2). But you're not done yet, since you haven't found the right and left bounds of the ellipse (ie, where the slope is undefined).

 

[physstudent1]

I'm confused though isn't a function not differentiable where it is undefined. Is it where you set the bottom equal to zero?

 

[atqamar]

To clarify that question, consider a circle: x^2 + y^2 = 2^2. Through implicit differentiation, you get \frac{dy}{dx}=-\frac{x}{y}. You know that (2,0) and (-2,0) are points on the circle, but if you differentiate at those points, you get \frac{dy}{dx}=\frac{2}{0},\frac{-2}{0}.

Similarly, an ellipse also has two points, the right and left bounds, which you are looking for. And the derivative is undefined. So yes, your approach of setting the denominator of \frac{dy}{dx} equal to zero, and then solving works fine.

 

[physstudent1]

great thanks for clearing that up the points I got for the box are(2,1)(-2,-1)(1,2)(-1,-2)

 

[atqamar]

Not so fast...

Your points are correct. But you must realize that this ellipse is not symmetric to the x and y axis. It is an oblique ellipse. Plot your points (the point (0,0) is the center), and draw the slopes, then extend the lines into an ellipse and you'll see that the four points are not the vertices of your rectangle. If you want the box to completely encapsulate the box, then you need to adjust the points. HINT: The box will be tangent to the four points you solved for.

 

[physstudent1]

oh, I see now that the points are not the vertices of the box but I am pretty lost on how to adjust them to get the vertices of the box I do understand the box must be tangent to these points but I don't know what to do from here

 

[atqamar]

Sorry, my explanation wasn't clear. You now know 4 points on the ellipse. You also know the derivative (slope) of those points.
These two points have slope of 0 (horizontal lines): (1,2)(-1,-2). So you can extend horizontal lines from those two points. And these two points have undefined slope (vertical lines): (2,1)(-2,-1). So you can extend vertical lines from those points. And voila, you should end up with your box. Finding the vertex of the box is easy from there.

 

[physstudent1]

ic, I understand now that extending these slopes brings the box but how do you get the actual points for the vertices?

EDIT: I graphed it and I see it now thank you.

 

[D H]

You got the horizontal lines by finding where dy/dx=0. To obtain the vertical lines, do the same for dx/dy.

 

[Smartass]

You got the horizontal lines by finding where dy/dx=0. To obtain the vertical lines, do the same for dx/dy.

Exactly! Put dx/dy=0 to get the vertical lines.

 

[physstudent1]

vertices of box (2,-2)(2,2)(-2,-2)(-2,2) dimensions 4 x 4 ?

 

[atqamar]

Perfect.