Tilted Ellipse Question?
You have a fixed circle O with a point P on the circumference. Point P is the center of circle P. Point Q is on the circumference of circle P. As point P moves around circle O, point Q traces out a curve. Circle P remains tangent to the line y = -a as point P moves around circle O. Circle O is tangent to the same line.
What is the equation and nature of the curve traced out by point Q? Prove or disprove that it is an ellipse. Please see the link for an illustration.
I came up with this completely by accident when I was working on something entirely different.
Yes, PQ is always a horizontal line.
The points (0,0); (2,0); and (0,-1) are on the curve.
On my last post I should have said, for a = 1 those points are on the curve.
In your second edit you are correct. It is a tilted ellipse and that is its equation. The area is also correct.
I am interested in how you derived the equation to compare it to my approach. I don't need to see all the number crunching but am interested in a brief description of the approach.
Thanks. I solved it using parametric equations. I noticed that y was part of the parametric equation for x. Then I converted to Cartesian.
- Scythian1950Lv 71 decade agoFavorite Answer
I assume that the slope of the line through P and Q remains constant? Otherwise, what constrains point Q?
It's a tilted ellipse, and the equation is:
x²(1 - 2Sin(A) + (Sin(A))²) + 2xy(Cos(A)(Sin(A) - 1)) + y²(1 + (Cos(A))²) + 2y(Sin(A) -1) = 0
where A is the angle line PQ makes with the horizontal. Note that it passes through (0,0).
Edit: Yes, if PQ is always horizontal, then it would be a special case of the more general ellipse as given above. The equation would reduce to:
x² - 2xy + 2y² - 2y = 0
Edit 2: Whoopsies! I'm using your (0, -a) as my (0, 0). My bad. But it doesn't change the fact it's an tilted ellipse. The equation should be revised to (for a = 1):
x² - 2xy + 2y² - 2x + 2y = 0
Bonus: This has an area of π.
Edit: I first worked out the coordinates x and y in terms of angle B which line OP makes with the horizontal. Frequently it's not readily possible to convert parametric equations into an implicit function f(x,y), but since I was able to express y in terms of Sin(B), it was just a matter of plugging that into the equation for x. In the simpler case where A = 0, you can guess that it's an affine transform of the circle O with the same area, because you can see that the horizontal width of the circle is preserved. However, the slope of the tilted ellipse is the unwieldy figure of:
rather than 1 as one would initially jump into that conclusion. The line connecting the bottom and top tangents of the ellipse is 1, however, as it should be.