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 $O A=\alpha$ and $O B=b$. 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 $A B$ and let the coordinates of $Q(h, k)$. Here, $a$ and $b$ are the variable and we have to find locus of $Q$.
Given,
$A B=c$
$\Rightarrow A B^{2} =c^{2} $
$\Rightarrow a^{2}+b^{2} =c^{2} ....$ (i)
Since, $P Q$ is perpendicular to $A B$.

$\Rightarrow$ Slope of $A B \cdot$ Slope of $P Q=-1$
$\Rightarrow \frac{0-b}{a-0} \cdot \frac{k-b}{h-a}=-1$
$\Rightarrow b k-b^{2}=a h-a^{2}$
$\Rightarrow a h-b k=a^{2}-b^{2}$
Equation of line $A B$ is $\frac{x}{a}+\frac{y}{b}=1$.
Since, $Q$ lies on $A B$, therefore
$\frac{h}{a}+\frac{k}{b}=1$
$\Rightarrow b h+a k=a b ....$(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 $x^{2 / 3}+y^{2 / 3}=c^{2 / 3}$.