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, the locus of the foot of the perpendicular drawn from $P$ to $AB$ is ?

• A. $x^{2/3}+y^{2/3}=c^{1/3}$
• B. $x^2+y^2=c^2$
• C. $x^{1/3}+y^{1/3}=c^{1/3}$
• D. $x^{2/3}+y^{2/3}=c^{2/3}$

Answer 1

The correct option is D $x^{2/3}+y^{2/3}=c^{2/3}$

Let $OA=a$ and $OB=b$. Then, the co-ordinates of $A$ and $B$ are $(a,0)$ and $(0,b)$ respectively.

The co-ordinates 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 $Q.$

Now, $AB=c$
$\Rightarrow A{B}^2=c^2\Rightarrow O{A}^2+O{B}^2=c^2$

$\Rightarrow a^2+b^2+c^2$ .......(1)

Now $PQ\perp AB$
$\Rightarrow$ Slope of $AB.$ Slope of $PQ=-1$

$\Rightarrow \frac{k-b}{h-a}\cdot \frac{0-b}{a-0}=-1$

$\Rightarrow bk-b^2=ah-a^2$

$\Rightarrow ah-bk=a^2-b^2$ .......(2)

Equation of line $AB$ is $\frac{x}{a}+\frac{y}{b}=1$

Since, $Q$ lies on $AB,$
$\frac{h}{a}+\frac{k}{b}=1\Rightarrow bh+ak=ab$ .......(3)

Solving (2) and (3), we get
$\frac{h}{a{b}^2+a(a^2-b^2)}=\frac{k}{-b(a^2-b^2)+a^{2}b}=\frac{1}{a^2+b{b}^2}$

$\Rightarrow \frac{h}{a^3}=\frac{k}{b^2}=\frac{1}{c^2}$ {using (1)}

$\Rightarrow a=(h{c}^2{)}^{1/3}$ and $b=(k{c}^2{)}^{1/3}$

Substituting the values of $a$ and $b$ in $a^2+b^2+c^2$, we get

$h^{2/3}+k^{2/3}=c^{2/3}$

$x^{2/3}+y^{2/3}=c^{2/3}$ required locus.