An ellipse is inscribed in Triangle ABC: $A(0,0), B(1,2)$ and $C(2,0)$. One of the focus of the ellipse is at S(1,$\frac {1}{2}$). Find the eccentricity $e$ of this ellipse.
I know this looks like an easy question but I don't know how to go about doing it. Any help will be appreciated.
As noted in the comment the ellipse has a symmetry axis $x=1$, so it's better to start with a translation of this axit to $x=0$ , that does not change the geometry of the figure. This means that now we have: $A=(-1,0)$, $B=(0,2)$, $C=(1,0)$ and $F=(0,1/2)$.
By the symmetry we know that the directrix $d$ has equation $y=-d$ and we can use the equation of an ellipse in the form $\overline{ XF}=e \overline{Xd}$ , where $e$ is the eccentricity, and $X$ is a generic point $X=(x,y)$.
So (substituting the given coordinates and squaring) we find the equation: $$ x^2+\left(y-\frac{1}{2}\right)^2=e^2(y+d)^2 $$
Since (always by symmetry) the point $(0,0)$ is a vertex of the ellipse, substituting its coordinates we find: $ed=\pm\frac{1}{2}$, so we have two ellipses (as we can aspect).
Now ve develop only the case $d=\frac{1}{2e}$ that gives, for the equation of the ellipse: $$ x^2+\left(y-\frac{1}{2}\right)^2=e^2\left( y+\frac{1}{2e}\right)^2 $$
We want that the ellipse be tangent to a side of the triangle (by symmetry it's tangent also to the other side), so we take the side $AB$ that stay on the line $y=2(x+1)$ or, better, $x=\frac{y}{2}-1$, substituting this value of $x$ in the equation of the ellipse, we have a second degree equation , and the line is tangent to the ellipse iff the discriminant$\Delta$ of this equation is null.
Solving the equation $\Delta=0$, that is an equation with unknown $e$ we find the value of the eccentricity. This require a bit of algebra, but it's not difficult and you find that, since we want $0<e<1$ (for an ellipse) the solution is $e=1/5$.
Using $ed=-\frac{1}{2}$ you can find another ellipse.