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 $OA=\alpha$ and $OB=b$. Then, the coordinates of $A$ and $B$ are $(a,0)$ and $(0,b)$ respectively and also, coordinates of $P$ are $(a,b)$. Let $\theta$ be the foot of perpendicular from $P$ on $AB$ and let the coordinates of $Q(h,k)$. Here, $a$ and $b$ are the variable and we have to find locus of $Q$.
Given,
$AB=c$
$\Rightarrow A{B}^2=c^2$
$\Rightarrow a^2+b^2=c^2....$ (i)
Since, $PQ$ is perpendicular to $AB$.

$\Rightarrow$ Slope of $AB\cdot$ Slope of $PQ=-1$
$\Rightarrow \frac{0-b}{a-0}\cdot \frac{k-b}{h-a}=-1$
$\Rightarrow bk-b^2=ah-a^2$
$\Rightarrow ah-bk=a^2-b^2$
Equation of line $AB$ is $\frac{x}{a}+\frac{y}{b}=1$.
Since, $Q$ lies on $AB$, therefore
$\frac{h}{a}+\frac{k}{b}=1$
$\Rightarrow bh+ak=ab....$(iii)
On solving Eqs. (ii) and (iii), we get
$\frac{h}{a{b}^2+a\left(a^2-b^2\right)}=\frac{k}{-b\left(a^2-b^2\right)+a^{2}b}=\frac{1}{a^2+b^2}$
$\Rightarrow \frac{h}{a^3}=\frac{k}{b^3}=\frac{1}{c^2}$ [from Eq. (i)]
$\Rightarrow a={\left(h{c}^2\right)}^{1/3}$ and $b={\left(k{c}^2\right)}^{1/3}$
On substituting the values of $a$ and $b$ in $a^2+b^2=c^2$,
we get $h^{2/3}+k^{2/3}=c^{2/3}$
Hence, locus of a point is $x^{2/3}+y^{2/3}=c^{2/3}$.