Question

The ends $A,B$ of a straight line segment of constant length $c$ slide upon the fixed rectangular axes $OX,OY$ respectively. If the rectangle $OAPB$ be completed, then show that the locus of the foot of the perpendicular drawn from $P$ to $AB$ is $x^{2/3}+y^{2/3}=c^{2/3}$

Answer 1

Let the coordinates of $A$ and $B$ be $(a,0)$ and $(0,b)$ on the axes so that its equation is $x/a+y/b=1$ or $bx+ay-ab=\mathrm{0......}\left(1\right)$
By given condition $AB=c$ or $a^2+b^2=c^2....\left(2\right)$
Again if $P$ be the vertex of rectangle then point $P$ is $(a,b)$
If $(h,k)$ be the foot of perpendicular from $P(a,b)$ on $AB$, then by rule
$\frac{h-a}{\frac{1}{a}}=\frac{k-b}{\frac{1}{b}}=-\frac{1+1-1}{\frac{1}{a^2}+\frac{1}{b^2}}=-\frac{a^2{b}^2}{a^2+b^2}$
$h=a-\frac{a^2{b}^2}{a^2+b^2}=\frac{a^2}{c^2}ora={\left(c^{2}h\right)}^{1/3}$
Similarly $b={\left(c^{2}k\right)}^{1/3}$ Putting in (2) and generalizing we get
$c^{4/3}(x^{2/3}+y^{2/3})=c^{2}or{x}^{2/3}+y^{2/3}=c^{2/3}$