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
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
Show Hint
For locus problems involving a midpoint, first express the endpoints in terms of the midpoint coordinates and then use the given geometric condition.
Updated On: Jun 18, 2026
- \(y=12x\)
- \[ \left(x-\frac{y}{3}\right)^2+y^2=\frac{r^2}{2} \]
- \[ \left(x-\frac{y}{3}\right)^2+y^2=\frac{r^2}{4} \]
- \(y=6x\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Take coordinates of the endpoints.
Let \[ A(a,0) \] be on the \(x\)-axis.
Since \(B\) lies on \[ y=6x, \] let \[ B(t,6t). \]
Step 2: Use the fixed length condition.
Given \[ AB=r. \] Therefore, \[ (a-t)^2+(0-6t)^2=r^2. \] \[ (a-t)^2+36t^2=r^2. \] \[ (a-t)^2+(6t)^2=r^2. \]
Step 3: Coordinates of the midpoint.
Let the midpoint be \[ M(x,y). \] Then \[ x=\frac{a+t}{2}, \qquad y=\frac{0+6t}{2}=3t. \] Hence, \[ t=\frac{y}{3}. \] Also, \[ a=2x-t = 2x-\frac{y}{3}. \]
Step 4: Substitute in the length equation.
From \[ (a-t)^2+36t^2=r^2, \] we get \[ \left(2x-\frac{y}{3}-\frac{y}{3}\right)^2 + 36\left(\frac{y}{3}\right)^2 = r^2. \] \[ \left(2x-\frac{2y}{3}\right)^2+4y^2=r^2. \] Dividing by \(4\), \[ \left(x-\frac{y}{3}\right)^2+y^2 = \frac{r^2}{4}. \]
Step 5: Final conclusion.
Hence the locus of the midpoint is \[ \boxed{\left(x-\frac{y}{3}\right)^2+y^2=\frac{r^2}{4}}. \]
Let \[ A(a,0) \] be on the \(x\)-axis.
Since \(B\) lies on \[ y=6x, \] let \[ B(t,6t). \]
Step 2: Use the fixed length condition.
Given \[ AB=r. \] Therefore, \[ (a-t)^2+(0-6t)^2=r^2. \] \[ (a-t)^2+36t^2=r^2. \] \[ (a-t)^2+(6t)^2=r^2. \]
Step 3: Coordinates of the midpoint.
Let the midpoint be \[ M(x,y). \] Then \[ x=\frac{a+t}{2}, \qquad y=\frac{0+6t}{2}=3t. \] Hence, \[ t=\frac{y}{3}. \] Also, \[ a=2x-t = 2x-\frac{y}{3}. \]
Step 4: Substitute in the length equation.
From \[ (a-t)^2+36t^2=r^2, \] we get \[ \left(2x-\frac{y}{3}-\frac{y}{3}\right)^2 + 36\left(\frac{y}{3}\right)^2 = r^2. \] \[ \left(2x-\frac{2y}{3}\right)^2+4y^2=r^2. \] Dividing by \(4\), \[ \left(x-\frac{y}{3}\right)^2+y^2 = \frac{r^2}{4}. \]
Step 5: Final conclusion.
Hence the locus of the midpoint is \[ \boxed{\left(x-\frac{y}{3}\right)^2+y^2=\frac{r^2}{4}}. \]
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