Question

If a line \(AB\) of length \(r\) moves such that \(A\) and \(B\) always lie respectively on the \(x\)-axis and the line \[ y=6x, \] then the locus of the midpoint of \(AB\) is

• A. \(y=12x\)
• B. \[ \left(x-\frac{y}{3}\right)^2+y^2=\frac{r^2}{2} \]
• C. \[ \left(x-\frac{y}{3}\right)^2+y^2=\frac{r^2}{4} \]
• D. \(y=6x\)

Hint:

For locus problems involving a midpoint, first express the endpoints in terms of the midpoint coordinates and then use the given geometric condition.

Answer 1

The Correct Option is C

Step 1: Assign coordinates to the endpoints.
Let A lie on the x-axis, so A = (a, 0). Let B lie on the line y = 6x, so B = (t, 6t).

Step 2: Express the fixed distance condition.
Given AB = r, the distance formula yields (a - t)² + (0 - 6t)² = r², which simplifies to (a - t)² + 36t² = r².

Step 3: Parametrize the midpoint M(x, y).
x = (a + t)/2 and y = (0 + 6t)/2 = 3t. Hence t = y/3 and a = 2x - t = 2x - y/3.

Step 4: Substitute the parameters into the distance equation.
(a - t)² + 36t² = r² becomes [2x - y/3 - y/3]² + 36(y/3)² = r², which simplifies to (2x - 2y/3)² + 4y² = r². Dividing by 4 gives (x - y/3)² + y² = r²/4.

Step 5: Final conclusion.
The locus of the midpoint is (x - y/3)² + y² = r²/4.