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.

Dahdahda Mar 20, 2019

#1

#8**+1 **

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

dierdurst Mar 20, 2019