SATHEE: Straight Line and Pair of Straight Lines 1 Question 66

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)

Show Answer

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 $θ$ 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,

$A B = c \Rightarrow A B^{2} = c^{2}$

$\Rightarrow O A^{2} + O B^{2} = c^{2} \Rightarrow a^{2} + b^{2} = c^{2}$

Since, $P Q$ is perpendicular to $A B$ .

$\Rightarrow$ Slope of $A B \cdot$ Slope of $P Q = - 1$

$\begin{aligned} \Rightarrow \frac{0 - b}{a - 0} \cdot \frac{k - b}{h - a} = - 1 \\ \Rightarrow b k - b^{2} = a h - a^{2} \\ \Rightarrow a h - b k = a^{2} - b^{2} \end{aligned}$

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

Since, $Q$ lies on $A B$ , therefore

$\begin{aligned} \frac{h}{a} + \frac{k}{b} & = 1 \\ \Rightarrow b h + a k & = a b \end{aligned}$

On solving Eqs. (ii) and (iii), we get

$\begin{aligned} \frac{h}{a b^{2} + a (a^{2} - b^{2})} = \frac{k}{- b (a^{2} - b^{2}) + a^{2} b} = \frac{1}{a^{2} + b^{2}} \\ \Rightarrow & \frac{h}{a^{3}} = \frac{k}{b^{3}} = \frac{1}{c^{2}} [from Eq. (i)] \\ \Rightarrow a = {(h c^{2})}^{1 / 3} & and b = {(k c^{2})}^{1 / 3} \end{aligned}$