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Here's the story:

I am a high school student who absolutely loves math. So I took a university level mathematics course that is renowned throughout our school for being extremely rigorous and tough. Last week we had our first lesson, and this quarter we'll be focusing on analytical geometry. Our teacher, known at our school for being the best mathematician, gave us this as homework: Find the cartesian equation of the ellipse

This probably is extremely easy for you guys, but don't forget this was our very first lesson. I couldn't solve it and today in class I found out nobody actually solved this. But what really shocked me, our teacher couldn't solve this either! He actually had no clue at all, he stared at it for like 20 minutes, and tried like FP=PQ but he couldn't solve it! So my question to you guys is, what is the solution (not only the answers but also explanation). I want to assure you guys that this is not our homework, and I want to know the answer as my 'smart' teacher doesn't know it either and I feel as if the only forum where people could answer these questions (with pleasure and) without a problem is this one.

PS - I am adding information as on the intial question there was a lot of uncertainty and I don't exactly know how this forum works (this is my first post), so here is some extra information:

  • M (-a,0) and F (a,0)
  • The ellipse is shown (approximately) by the ellipse in red, and, this is important: is a figure of all the points which are as far from the circle as from F (I'm Dutch, so the english terminalogy is tough, but I hope you know what I mean).
  • r is the radius
  • P (x,y)
  • And there were 3 questions (but because he didn't even know the first one the 2 others were completely disregarded but these are the 3 questions):

    1. What is the cartesian equation (I believe it is called that) of the 'ellipse'
    2. What kind of figure would you get if you place F outside of the circle ( I believe he briefly said it would be a hyperbole but I'm not sure)
    3. And what would be the equation of that figure (the hyperbole).
  • I'm extremely sorry for my English, I'm just so desperate to know the answer and not a single mathematics teacher at our school knows this.. it's a shame.

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Without further info I'm afraid it is pretty impossible to solve this: (1) Is M that circle's center? Is F a point on the x-axis s.t. $\,|M|=|F|\,$ If this is so, it then must be that $\,r>2a\,$...Assuming all this, is then that ellipse the one that has $\,M\,,\,F\,$ as foci and, thus, its main radius is on the x-axis? – DonAntonio Sep 7 '12 at 18:13
dear LineofView, could you tell us the exact assignment? What are the points and what exactly should be an ellipse? – Olivier Bégassat Sep 7 '12 at 20:00
You needn’t apologize for your English: it’s very good. – Brian M. Scott Sep 7 '12 at 20:54
You should have edited your previous question instead of making a new one. Please do so next time. – Henning Makholm Sep 8 '12 at 1:37

It's hard to tell from the picture exactly what you're using as the definition of the ellipse. One way of defining an ellipse is that the sum of distances from the foci to any point on the ellipse is constant. So let's say the foci are at $(a,0)$ and $(-a,0)$ and the sum of the distances is $d$. Then the point $(x,y)$ is on the ellipse if $$ \sqrt{(x-a)^2 + y^2} + \sqrt{(x+a)^2 + y^2} = d $$ Move the second square root to the right side, square both sides, and expand. You get $$ (x-a)^2 + y^2 = (x+a)^2 + y^2 - 2 d \sqrt{(x+a)^2 + y^2} + d^2 $$ After some more expansion, cancellation, and moving terms around, $$ 2 d \sqrt{(x+a)^2 + y^2} = d^2 + 4 a x $$ Again square both sides and expand: $$ 4\,{d}^{2}{x}^{2}+8\,{d}^{2}ax+4\,{a}^{2}{d}^{2}+4\,{d}^{2}{y}^{2}=16 \,{x}^{2}{a}^{2}+8\,{d}^{2}ax+{d}^{4}$$ Cancel the $8 d^2 a x$'s, bring the terms containing $x^2$ and $y^2$ to the left and the constant $4 a^2 d^2$ to the right, and collect terms: $$ (4 d^2 - 16 a^2) x^2 + 4 d^2 y^2 = d^4 - 4 a^2 d^2 $$

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The only thing the teacher said that the answer should be something like (I forgot exactly what) x^2/a^2 + ... = r^2 but this is not the good answer I only remember him saying something in the general area of this – LineofView Sep 7 '12 at 18:40
Divide by the right side to get it in the form $x^2/A^2 + y^2/B^2 = 1$. – Robert Israel Sep 7 '12 at 22:06

It should be immediately clear that this is an ellipse if you have already proved that the ellipse can be defined as the set of all points $(x,y)$ such that the sum of the distances from $(x,y)$ to the Foci (in this case represented by $(0,a)$ and $(0,-a)$) is constant. This is often taken to be the very definition of an ellipse. Can you see from your figure why this must be the case? Once you have proved that this is the case for your figure, you are done. If you are familiar with norms, you can write this as follows: (try to do it yourself before peeking)

$\lVert (x,y)-(0,a)\rVert + \lVert(x,y)-(0,-a)\rVert = r$

Unwrapping this equation gives you the Cartesian equation for the ellipse in terms of $a$ and $r$. It should be clear how to come to a similar conclusion if $F$ is placed outside the circle by remembering a fact about the hyperbola - the hyperbola is all points $(x,y)$ such that the distance from $(x,y)$ to one of the foci, say $(0,-a)$ minus the distance from $(x,y)$ to the other foci is a constant (which will precisely be the case when you move $F$ outside the circle.

Note: If, instead, you only started with the Cartesian equation of an ellipse as the definition, which is sometimes done, you can still solve the problem the same exact way. The information in the picture yields precisely the fact that the points included are those which satisfy the above equation, and some algebra will show that the above equation is indeed the Cartesian equation for an ellipse.

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Here's a hint for part 1:

For an ellipse, if you knew the length of the semi-major and semi-minor axes, you'd be done, right? (Then the equation is $x^2 / L_1^2 + y^2 / L_2^2 = 1$).

For the length of the semi-major axis (the horizontal one), just draw the line from M through F to the edge of the circle, and you should be able to set up a simple equation that solves for what you need (your answer will involve $a$ and $r$, of course).

For the semi-minor axis, draw a picture something like this: enter image description here

You want to solve for length of $ON$, and you know that:

  • The red and green lines combined are of length $r$
  • The green and blue line are the same length
  • Side MO and FO are both length $a$
  • Side ON is congruent to itself (do you see where I'm going yet...?)
  • $\angle MON$ and $\angle FON$ are both right angles

You should be able to work out the rest.

You may additionally need to check that this figure is actually an ellipse. One way to handle this is to proceed as I suggest above, then verify this shape satisfies a more standard definition of ellipses - in particular, once the length of the semi-major axis is known, it's straightforward to verify "The sum of the distances from any point P on the ellipse to those two foci is constant and equal to the length of the major axis (i.e. $2 \times$ semi-major axis)"

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Assuming the shape that has been described really is an ellipse, then knowing the length of the major and minor radii would give you an ellipse equation. But I am wondering if OP can even assume ahead of time that this is a true ellipse. The shape is defined in a way that ellipses are not usually defined, and I wonder if OP is looking to directly derive that an equation for the shape can be a standard Cartesian ellipse equation. – alex.jordan Sep 7 '12 at 21:43
That's a fair objection. I now suggest a way of checking it is, in fact, an ellipse. – BaronVT Sep 7 '12 at 22:05

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