Straight Line and Pair of Straight Lines 1 Question 66
66. The ends $A, B$ of a straight line segment of constant length $c$ slide upon the fixed rectangular axes $O X, O Y$ respectively. If the rectangle $O A P B$ be completed, then show that the locus of the foot of the perpendicular drawn from $P$ to $A B$ is
$$ x^{2 / 3}+y^{2 / 3}=c^{2 / 3} $$
(1983, 2M)
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Answer:
Correct Answer: 66. $c=-4,(4,4),(2,0)$
Solution:
- Let $O A=a$ 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,
$$ A B=c \Rightarrow A B^{2}=c^{2} $$
$\Rightarrow \quad O A^{2}+O B^{2}=c^{2} \Rightarrow a^{2}+b^{2}=c^{2}$
Since, $P Q$ is perpendicular to $A B$.
$\Rightarrow$ Slope of $A B \cdot$ Slope of $P Q=-1$
$$ \begin{aligned} & \Rightarrow \quad \frac{0-b}{a-0} \cdot \frac{k-b}{h-a}=-1 \\ & \Rightarrow \quad b k-b^{2}=a h-a^{2} \\ & \Rightarrow \quad a h-b k=a^{2}-b^{2} \end{aligned} $$
Equation of line $A B$ is $\frac{x}{a}+\frac{y}{b}=1$.
Since, $Q$ lies on $A B$, therefore
$$ \begin{array}{rlrl} & & \frac{h}{a}+\frac{k}{b} & =1 \\ \Rightarrow \quad b h+a k & =a b \end{array} $$
On solving Eqs. (ii) and (iii), we get
$$ \begin{aligned} & \frac{h}{a b^{2}+a\left(a^{2}-b^{2}\right)}=\frac{k}{-b\left(a^{2}-b^{2}\right)+a^{2} b}=\frac{1}{a^{2}+b^{2}} \\ \Rightarrow \quad & \frac{h}{a^{3}}=\frac{k}{b^{3}}=\frac{1}{c^{2}} \quad \text { [from Eq. (i)] } \\ \Rightarrow \quad a=\left(h c^{2}\right)^{1 / 3} & \text { and } \quad b=\left(k c^{2}\right)^{1 / 3} \end{aligned} $$
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 $x^{2 / 3}+y^{2 / 3}=c^{2 / 3}$.
