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We have an ellipse with a circle in it. The circle is passing through the two vertices and through the ellipse's center. It's diameter equals 7. We have also an equilateral triangle which vertices are in ellipse's focuses and in the minor vertex (the triangle's height is equal to semi-minor axis).

How can we define the canonical equation of the ellipse in this case?

plot

Ferra
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ivkremer
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2 Answers2

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The sum of the distance between one point of the ellipse and the two vertices is constant.

So the major axis length is $2d$, $d$ being the distance between the two vertices of the ellipse. Let's call $k$ half the minor axis of the ellipse.

We deduce from the diameter of the circle that:

$k²+d²=7²$, that is $d²=7²-k²$

$k²+(\frac{d}{2})²=d²$, that is $k²=\frac{3d²}{4}$

Thus, $d²=28$

The equation of the ellipse is then

$\dfrac{X²}{28}+\dfrac{Y²}{21}=1$

Martigan
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  • Thank you very much. But am I right that you denoted the semi-major axis as d? I'm asking that because you said major axis length is 2d, but d is the distance between the two vertices of the ellipse. I think you meant the distance between the vertex and the center, didn't you? – ivkremer Oct 20 '14 at 12:19
  • And the second question is how did you know that k^2 + (d/2)^2 = d^2? – ivkremer Oct 20 '14 at 12:21
  • Both the distance between the two vertices and semi-major axis are of length $d$. – Martigan Oct 20 '14 at 13:41
  • @Kremchik Look at your drawing. The triangle $OF_{2}V_{2}$ (O being the origin and V2 the second vertice (vertical one) is a right angle triangle, with size length $d$ ($F_{2}V_{2}$) and $d/2$ ($OF_{2}$), since $O$ is the middle of $F_{1}F_{2}$. – Martigan Oct 20 '14 at 14:03
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Hint:

Starting with $F_{2}=\left\langle c,0\right\rangle $ and denoting the intersection of the ellips, circle and $x$-axis by $V$ we find:

$$\left\Vert V-F_{1}\right\Vert +\left\Vert V-F_{2}\right\Vert =4c$$ and consequently:

$$E:=\left\{ \left\langle x,y\right\rangle :\left\Vert \left\langle x,y\right\rangle -\left\langle c,0\right\rangle \right\Vert +\left\Vert \left\langle x,y\right\rangle -\left\langle -c,0\right\rangle \right\Vert =4c\right\} =\left\{ \left\langle x,y\right\rangle :3x^{2}+4y^{2}=12c^{2}\right\}$$

drhab
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