Solution .

For convenience, since segment $AC$ is divided into four equal parts, we introduce coordinates scaled so that $AC=4$ . We place the origin at the center of the (semi)ellipse, which is to say the midpoint between the foci $F$ and $G$ . Thus the equation of the ellipse will have the standard form ${x}^{2}/{a}^{2}+{y}^{2}/{b}^{2}=1$ for some constants $a$ and $b$ giving the positive $x$ – and $y$ -intercepts of the ellipse. We’ve included this coordinate system in Figure 2.

Figure 2: Diagram for Semi Inellipse with coordinate system imposed and auxiliary point $D$ .

The coordinates of some of the points are immediate:$$A(-\frac{5}{2},0),{\textstyle \phantom{\rule{0.167em}{0ex}}}I(-\frac{3}{2},0),{\textstyle \phantom{\rule{0.167em}{0ex}}}F(-\frac{1}{2},0),{\textstyle \phantom{\rule{0.167em}{0ex}}}G(\frac{1}{2},0),{\textstyle \phantom{\rule{0.278em}{0ex}}}\text{and}{\textstyle \phantom{\rule{0.278em}{0ex}}}C(\frac{3}{2},0).$$ Since $C$ is the $x$ -intercept of the ellipse, we have $a=3/2$ . We also added the $y$ -intercept $D(0,b)$ in Figure 2. Since $FC+CG=3$ , we have by the definition of an ellipse that also $3=FD+DG=2\sqrt{{b}^{2}+1{\textstyle \phantom{\rule{-0.167em}{0ex}}}{\textstyle \phantom{\rule{-0.167em}{0ex}}}/{\textstyle \phantom{\rule{-0.167em}{0ex}}}{\textstyle \phantom{\rule{-0.167em}{0ex}}}4}$ , from which we solve for $b=\sqrt{2}$ .

Now suppose the coordinates of point $B$ are $(3\text{/}2,h)$ . Then the equation of line $AB$ is$$y=\frac{h}{4}(x+\frac{5}{2}).$$ Substituting this value for $y$ into the equation of the ellipse to solve for the $x$ -coordinate of point $T$ (since the equations of both the line and the ellipse must be satisfied at point $T$ ), we get

We can write out the rightmost square and multiply both sides by $1152=9\cdot 32\cdot 4$ to obtain the quadratic However, it is given in the problem that the line $AB$ is tangent to the semiellipse, so they only intersect in one point. Therefore this quadratic must have only one real root. But by the quadratic formula for $A{x}^{2}+Bx+C=0$ , namely $x=(-B\pm \sqrt{{B}^{2}-4AC})/2A$ , there are two roots unless ${B}^{2}-4AC=0$ . In other words, it must be the case that Again multiplying out, this equation simplifies to ${h}^{2}=8$ , amazingly enough. Armed with this information, it is easy to compute that $AB=\sqrt{16+{h}^{2}}=2\sqrt{6},$ so $AB/AC=\sqrt{6}/2$ .