I think I may know what you mean. If so, just sub in your values for what I have in my example.
Suppose you have a road going through a semi-elliptical tunnel(like in the diagram).
The road is 30 feet wide and the highest part of the arch is 10 feet.
Say, we wanted to find the height of the arch 6 feet from the center of the road.
Use the general ellipse equation formula \(\displaystyle \L\\\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
In this example, a=15(major axis, the road) and b=10(minor axis, the height at center).
We could write this as \(\displaystyle V(\pm{15},0)\;\ M(0, \pm{10})\)
Substitute x=6: \(\displaystyle \L\\\frac{6^{2}}{15^{2}}+\frac{y^{2}}{10^{2}}=1\)
=\(\displaystyle \L\\\frac{36}{225}+\frac{y^{2}}{100}=1\)
Now, solve for y:
\(\displaystyle \L\\\frac{y^{2}}{100}=1-\frac{36}{225}\)
\(\displaystyle \L\\=\frac{y^{2}}{100}=\frac{21}{25}\)
\(\displaystyle \L\\y^{2}=\frac{2100}{25}=84\)
\(\displaystyle \L\\y=\sqrt{84}=2\sqrt{21}\approx{9.165}\;\ ft\)
There, the height of your tunnel 6 feet on either side of the center of the road is about
9 feet 2 inches.
Now, if you wanted to pull a tape from the center of the road to that point on the arch we just calculated, you could do the Pythagoras thing.
\(\displaystyle \L\\\sqrt{6^{2}+84}=2\sqrt{30}=\approx{10.95}\;\ ft\)
About
10' 11-3/8"
I hope this is on the lines of your query.