Question

A fixed rod of length 5 units sliding between coordinates axes then find equation of locus of point which divides rod in the ratio 3 : 4.

• A. $\frac{x^2}{20^2} + \frac{y^2}{15^2} = \frac{1}{49}$
• B. $\frac{x^2}{15^2} + \frac{y^2}{20^2} = \frac{1}{49}$
• C. $\frac{x^2}{20^2} - \frac{y^2}{15^2} = \frac{1}{49}$
• D. $\frac{x^2}{15^2} - \frac{y^2}{20^2} = \frac{1}{49}$

Answer 1

Answer: (A)

$\frac{x^2}{20^2} + \frac{y^2}{15^2} = \frac{1}{49}$

Answer 2

Let the rod have endpoints A on the x-axis and B on the y-axis. Let A be $(a, 0)$ and B be $(0, b)$. The length of the rod is given as 5 units. Using the distance formula, the constraint is $a^2 + b^2 = 5^2 = 25$.

Let P be the point $(x, y)$ that divides the rod AB in the ratio 3:4. Using the section formula for internal division: $x = \frac{4 \cdot a + 3 \cdot 0}{3+4} = \frac{4a}{7}$ $y = \frac{4 \cdot 0 + 3 \cdot b}{3+4} = \frac{3b}{7}$

From these equations, we express $a$ and $b$ in terms of $x$ and $y$: $a = \frac{7x}{4}$ $b = \frac{7y}{3}$

Substitute these expressions for $a$ and $b$ into the constraint equation $a^2 + b^2 = 25$: $\left(\frac{7x}{4}\right)^2 + \left(\frac{7y}{3}\right)^2 = 25$ $\frac{49x^2}{16} + \frac{49y^2}{9} = 25$ Divide both sides by 25: $\frac{49x^2}{16 \cdot 25} + \frac{49y^2}{9 \cdot 25} = 1$ $\frac{x^2}{\frac{16 \cdot 25}{49}} + \frac{y^2}{\frac{9 \cdot 25}{49}} = 1$ $\frac{x^2}{\left(\frac{20}{7}\right)^2} + \frac{y^2}{\left(\frac{15}{7}\right)^2} = 1$ This equation is equivalent to: $\frac{x^2}{400/49} + \frac{y^2}{225/49} = 1$ Multiplying both sides by 49 gives: $\frac{49x^2}{400} + \frac{49y^2}{225} = 1$ This matches the given answer $\frac{x^2}{20^2} + \frac{y^2}{15^2} = \frac{1}{49}$.