See Fig. Let A = (a, 0) and B = (0, b) so that the equation of the line AB is
x/a + y/b = 1 ...(1)
and also
a2 + b2 = c2 ...(2)
Let M(x1, y1) be the foot of the perpendicular drawn from P onto the line AB. Since M(x1, y1) lies on AB, we have
x1/a + y1/b = 1 ....(3)
Since P = (a, b) and PM is perpendicular to AB, we have
Slope of Slope of PM x AB = -1
Therefore,
(b -y1/a - x1)(b - 0/0 - a) = 1
ax1 - by1 = a2 - b2 .....(4)
From Eqs. (3) and (4), we get
Therefore, the locus of M(x1, y1) is
x2/3 + y2/3 = c2/3