A rod of length `k` slides in a vertical plane, its ends touching the coordinate axes. Prove that the locus of the foot of the perpendicular from the origin to the rod is `(x^2+y^2)^3=k^2x^2y^2dot`

Explanation

(Slope of OP). (Slope of BP)` =-1 `
or `(beta)/(alpha).(beta-b)/(alpha-0).=-1`
or `b=(alpha^2+beta^2)/(beta)`
and (Slope of OP).(Slope of AP)`=-1`
or `(beta)/(alpha).(beta)/(alpha-alpha).=-1`
`therefore a=(alpha^2+beta^2)/(alpha)`
Now, `a^2+b^2=k^2`
`therefore (a^2_beta^2)^3=k^2a^2beta^2`
`therefore (x^2+y^2)^3=k^2x^2y^2`