A straight line segment of length l...

MathematicsDistance and Section FormulasJEE Advanced 1978Moderate

A straight line segment of length $l$ moves with its ends on two mutually perpendicular lines. Find the locus of the point which divides the line segment in the ratio $1 : 2$ .

Answer:9x^2 + 36y^2 = 4l^2

Visualized Solution (Hindi)

Visualizing the Moving Segment

Let the two perpendicular lines be the coordinate axes $O X$ and $O Y$ .
A line segment $A B$ of constant length $l$ moves such that $A$ is on the $x$ -axis and $B$ is on the $y$ -axis.

Defining Coordinates of Ends .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Let the coordinates of the ends be $A (a, 0)$ and $B (0, b)$ .
Here, $a$ and $b$ are the intercepts on the axes which change as the rod moves.

The Length Constraint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l

By the distance formula (or Pythagoras theorem) in $△ O A B$ :
$a^{2} + b^{2} = l^{2}$

Introducing Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (x, y)

Let $P (x, y)$ be the point dividing $A B$ in the ratio $1 : 2$ .
Using the Section Formula for internal division between $A (a, 0)$ and $B (0, b)$ with ratio $m : n = 1 : 2$ .

Section Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

For the $x$ -coordinate:
$x = \frac{1 ( 0 ) + 2 ( a )}{1 + 2}$

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

$x = \frac{2 a}{3}$
Rearranging for $a$ : $a = \frac{3 x}{2}$

Section Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

For the $y$ -coordinate:
$y = \frac{1 ( b ) + 2 ( 0 )}{1 + 2}$

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

$y = \frac{b}{3}$
Rearranging for $b$ : $b = 3 y$

Substituting into the Constraint

Substitute $a = \frac{3 x}{2}$ and $b = 3 y$ into $a^{2} + b^{2} = l^{2}$ :
$(\frac{3 x}{2})^{2} + (3 y)^{2} = l^{2}$

Squaring the Terms

Expanding the squares:
$\frac{9 x ^{2}}{4} + 9 y^{2} = l^{2}$

Final Locus Equation

Multiply the entire equation by $4$ to clear the fraction:
$9 x^{2} + 36 y^{2} = 4 l^{2}$
This is the equation of an ellipse.

The Way Forward

Key Takeaway: The locus of a point dividing a sliding segment in a fixed ratio is generally an ellipse.
Challenge: Find the locus if the point $P$ is the midpoint of $A B$ (ratio $1 : 1$ ). Does it become a circle?

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