[Geometry]
Problem: An ellipse in the first quadrant is tangent to both the $x$-axis and $y$-axis. One focus is at $(3,7)$, and the other focus is at $(d, 7)$. Compute $d$.

Solution: See the diagram below. The center of the ellipse is $C=\left(\frac{d+3}{2}, 7\right)$. The major axis of the ellipse is the line $y=7$, and the minor axis is the line $x=\frac{d+3}{2}$. The ellipse is tangent to the coordinate axes at $T_{x}=\left(\frac{d+3}{2}, 0\right)$ and $T_{y}=(0,7)$. Let $F_{1}=(3,7)$ and $F_{2}=(d, 7)$. Using the locus definition of an ellipse, we have $F_{1} T_{x}+F_{2} T_{x}=F_{1} T_{y}+F_{2} T_{y}$; that is,

$$
2 \sqrt{\left(\frac{d-3}{2}\right)^{2}+7^{2}}=d+3 \quad \text { or } \quad \sqrt{(d-3)^{2}+14^{2}}=d+3
$$

Squaring both sides of the last equation gives $d^{2}-6 d+205=d^{2}+6 d+9$ or $196=12 d$, so $d=\frac{49}{3}$.

<img_4034>

Final answer: $\frac{49}{3}$