The vertices are Q and U, P is a covertex. The foci (focal points) are R and T. The center is S. SQ = SU = a SP = b SR = ST = c By using the Pythagorean theorem on right triangle PST, we have STē + PSē = PTē cē + bē = PTē cē = PTē - bē If we can show that PT = a we will have shown what we needed to. By definition, every ellipse is such that any two points on the ellipse are such that the sum of the two distances from one of them to the two focal points (foci) equals the sum of the two distances from the other point to the two focal points (foci). P and U are two points on the ellipse. So by the definition of an ellipse, PR + PT = UR + UT Since PR = PT and UT = QR, PT + PT = UR + UT 2PT = UR + QR 2PT = QU Since SQ = SU = a, then QU = SQ + SU = a + a = 2a 2PT = 2a PT = a That was what we needed to show, since we have shown that cē = PTē - bē. So now we can substitute "a" for PT, and have what we wanted to show: cē = aē - bē Edwin