Problem above, from what I understand you are to find the vertical distance to an edge that is 6 feet from a vertex.

I'm trying to triangulate from the focus distances to the point to create an equation I can equate.

Using distance from the focus(c=$\displaystyle \sqrt{(25^2-14^2)}=20.71$

Setting the center at the origin, the distance from 2 foci = $\displaystyle 2(\sqrt{20.71^2+14^2}=50$

Therefore the point 6 feet from the vertex, which is 19 ft from the origin, and is 39.71ft from the last focus, is equated by:$\displaystyle 50-(39.71^2+b^2=c^2)$

This is where my approach halts, is my approach wrong? Or is there something I am just not seeing?

Edit answer is ~9.1