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.
If the first foci is \((3,7) \) and the second is \((d, 7)\), then \(C = \) \((\frac{d+3}{2}, 7)\). \(C\) is the center of the ellipse. The point where the ellipse touches the x-axis is \((\frac{d+3}{2}, 0)\).
We know that \(P\)(any points on the ellipse) = \(T\), so
\(2 \sqrt{\left( \frac{d - 3}{2} \right)^2 + 7^2} = d + 3\)
\(\sqrt{\left( \frac{d - 3}{2} \right)^2(4) + 7^2(4)} = d + 3\)
\(\sqrt{(d - 3)^2 + 196} = d + 3\)
Then, by squaring both sides, we get
\((d - 3)^2 + 196 = d^2 + 6d + 9\)
\(d^2-6d+9+196=d^2+6d+9\)
\(12d=196\)
\(d=\frac{49}{3}\)
- Daisy