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There are two lines, parallel to the $x$-axis, which pass through the foci and intersect the ellipse at four points. How can I find the points of intersection?

  • vertex: $(0,0)$
  • foci: $(0,10)$ and $(0,-10)$
  • co-vertices: $(20\sqrt2, 0)$ and $(-20\sqrt2, 0)$
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Find the equation of the ellipse first, then solve directly for the intersections. –  Potato Aug 26 '12 at 3:11
    
I've changed algebra tag to algebra-precalculus, since we don't use algebra tag anymore, see meta for details. –  Martin Sleziak Aug 26 '12 at 3:19

1 Answer 1

up vote 3 down vote accepted

You do not even need the equation for the ellipse if you use the property that, for any point $(x,y)$ on the ellipse, the sum of the distances from $(x,y)$ to the foci is a constant. Since we know that $(20\sqrt2,0)$ is a point on the ellipse, we can find this constant:

$\lVert (20\sqrt 2, -10)\rVert+\lVert(20\sqrt 2, 10)\rVert=2\sqrt{ 800+100}=60$.

Now we know that the y-coordinate of the points of intersection for the line going through $(0,10)$ is $10$, so we just need the x-coordinate. So we have that $$\lVert(x,10)-(0,10)\rVert + \lVert(x,10)-(0,-10)\rVert = 60\\ \lvert x \vert + \sqrt {x^2+400}=60\\ x^2+400=60^2-120\lvert x\rvert+x^2\\ \lvert x \rvert=\frac {3200}{120}=\frac {80} 3$$

So the points of intersection are $(\frac {80} 3 , 10)$ and $(\frac{-80} 3,10)$. The same method will allow you to find the other two points, try it! (You could also appeal to symmetry, but it's good to work through the argument on your own as well.)

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