# Thread: something to do with derivatives?

1. ## something to do with derivatives?

Can someone please help me out with this problem? I don't even know how to approach it. Thankyou

The graph of the tilted ellipse x^2 - x*y + y^2 = 3 is shown above. 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 ellipse.

2. Originally Posted by coe236
Can someone please help me out with this problem? I don't even know how to approach it. Thankyou

The graph of the tilted ellipse x^2 - x*y + y^2 = 3 is shown above. 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 ellipse.
the note says it all. the line you see are the vertical and horizontal tangent lines of the figure. so find the derivative and set it equal to zero to find the horizontal tangent lines, then make its denominator zero to find the vertical ones.

you do know how to find the derivative here, right? differentiate implicitly

3. Originally Posted by coe236
Can someone please help me out with this problem? I don't even know how to approach it. Thankyou

The graph of the tilted ellipse x^2 - x*y + y^2 = 3 is shown above. 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 ellipse.
Hello,

consider a horizontal tangent to the ellipse. This tangent has the equation

t: y = k (k is a constant)

Calculate the intersection between the ellipse and the tangent: You must get only one single point:

$\displaystyle x^2-k\cdot x + k^2 - 3=0~\implies~x=\frac12 \cdot k \pm \sqrt{\frac14 \cdot k^2 - k^2 + 3}$

You get exactly one point if the radical is zero:

$\displaystyle \frac14 \cdot k^2 - k^2 + 3~\implies~k=2\ \vee\ k= -2$

The upper and lower side of the square are placed on the lines with the equations: y = 2 or y = -2

To get the vertical lines you use x = r. All calculations are nearly the same as above. You'll get x = 2 or x = -2

4. Thankyou, I appreciate it