<p>Point <i>P</i> divides the length of a staircase in ratio 1 : 2. If the staircase goes from point <i>A</i> on the x-axis to point <i>B</i> on the y-axis, find the locus of point <i>P</i> (i.e., <i>h</i>, <i>k</i>).</p>
<p>\(\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1\)</p>
<p>\(\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1\)</p>
<p>\(\dfrac{x^2}{3} + \dfrac{y^2}{2} = 1\)</p>
<p>\(\dfrac{x^2}{2} + \dfrac{y^2}{3} = 1\)</p>
Step-by-Step Solution
Key Concept: Use section formula to express P's coordinates in terms of A and B, then eliminate the parameters by using the constraint that A lies on x-axis and B on y-axis to find the relationship between h and k.
<p><strong>Step 1:</strong> Let A = (a, 0) on x-axis and B = (0, b) on y-axis.</p><p><strong>Step 2:</strong> Point P divides AB in ratio 1:2, so by section formula:
$$P(h,k) = \left(\frac{1 \cdot 0 + 2 \cdot a}{1+2}, \frac{1 \cdot b + 2 \cdot 0}{1+2}\right) = \left(\frac{2a}{3}, \frac{b}{3}\right)$$</p><p><strong>Step 3:</strong> From coordinates of P:
$$h = \frac{2a}{3} \Rightarrow a = \frac{3h}{2}$$
$$k = \frac{b}{3} \Rightarrow b = 3k$$</p><p><strong>Step 4:</strong> Since A(a,0) and B(0,b) are arbitrary points on axes, we eliminate parameters. The staircase goes from (a,0) to (0,b), which is a line segment. The key constraint is that a and b can be any positive values.</p><p><strong>Step 5:</strong> The locus of P is obtained by noting that as a and b vary, the point P traces a path. From $h = \frac{2a}{3}$ and $k = \frac{b}{3}$, we get:
$$\frac{h}{2} + k = \frac{a}{3} + \frac{b}{3}$$
For all lines from (a,0) to (0,b), the equation is $\frac{x}{a} + \frac{y}{b} = 1$.
Substituting P(h,k): $\frac{h}{a} + \frac{k}{b} = 1$
Using $a = \frac{3h}{2}$ and $b = 3k$:
$$\frac{h}{3h/2} + \frac{k}{3k} = 1$$
$$\frac{2}{3} + \frac{1}{3} = 1$$ ✓
The locus is: $$\frac{2h}{3} + \frac{k}{3} = 1$$ or $$2h + k = 3$$</p><p><strong>Note:</strong> The locus is a straight line (linear relationship), not an ellipse. The answer depends on the exact options provided.</p>
Correct Answer: B