Problem 1-2 Ellipse Tracing

Problem Posts

Problem 1-2 (Geometry) [Difficulty 3] [PDF]

QUESTION
(Original) Define points $A\left(2,5+\sqrt{21}\right)$ , $B\left(2,5-\sqrt{21}\right)$ . Now let a point $P$ be initially at $P_{0}\left(4,5\right)$ . Let $Q$ be the point on ray $\overrightarrow{AP}$  such that $PB=PQ$ . As $P$ moves along the lower half of the ellipse $25x^{2}-100x+4y^{2}-40y+100=0$ (to eventually stop at $\left(0,5\right)$), the point $Q$  traces a path. Find the length of this path.

Overview/Laconic Solution Sketch
Apply geometric ellipse definition, and from that calculate arc length.



Discussion
The good news with problems of this level is when they ask about path length, the path in question is usually either a line or a circular arc. This is primarily because of the limitations of what arc length formulae are given in high school (for example, it's unlikely they require knowledge of elliptic integrals or other fancy arc lengths.)

The bad news is this is a geometry problem about an ellipse, which, compared to the usual fare, is a lumbering monster¹. That is, unless we consider a definition different from standard $\left(x-h\right)^2/a^2 + \left(y-k\right)^2/b^2=1$:

In mathematics, an ellipse is a curve on a plane surrounding two focal points such that a straight line drawn from one of the focal points to any point on the curve and then back to the other focal point has the same length for every point on the curve (Wikipedia)

You'll find this definition won't break down even if you subject the ellipse to any sort of rigid transformation. The one we're so used to is for the "tame" ellipses with axes along the coordinate axes. But I digress. Suppose a tame ellipse is longer horizontally (i.e. $a>b$).  Then, the foci are the points $\left(h\pm\sqrt{a^2-b^2},k\right)$ and the constant length referred to is $2a$.²

$$25x^{2}-100x+4y^{2}-40y+100=0\Rightarrow \frac{\left(x-2\right)^2}{4} + \frac{4\left(y-5\right)^2}{25} = 1$$

Now you'll immediately find from this that $A$ and $B$ are incidentally the foci of the ellipse. At this point, things appear contrived, and you know you're close to finishing the problem off. It now becomes evident that $AQ=AP+PQ=AP+PB=10$, so as we guessed, $Q$ traces a circular arc centered at $A$!

But what's the angular measure of that arc? Now we know that should be $2 \angle P_0AB$ (hint: use symmetry). Now $P_0A=5$ and we know $P_0$ has distance $2$ from $AB$ (semi-minor axis). So our angle is $2\sin^{-1}\frac{2}{5}$. Multiply that with $10$ for the answer.

Now, when written in "solutionese", all that condenses into this:

Solution [show]

$A$ and $B$ are the foci of the ellipse that $P$ must be on. This means that $AQ=AP+PQ=AP+PB$ is invariant. $Q$ then, traces out a circular arc. Now the major axis of the ellipse has length $10$. Furthermore, the angle that $AP$ initially makes with the vertical is $\sin^{-1}\left(\frac{2}{5}\right)$ , hence the arc length must be $\boxed{20\sin^{-1}\frac{2}{5}}$ units.

¹I formulated this problem to address a perceived lack of problems that involve conic section geometry. There are a lot of nice properties that the conic sections have, but unfortunately most conic problems only ask for center, axes and the like. Not this one.
²You can back-engineer this from the geometric definition using the Distance Formula.