Get canonical equation of ellipse

Question

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?

Answer 1

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

$\dfrac{X²}{28}+\dfrac{Y²}{21}=1$

Answer 2

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:

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