Let \(A\) be the point \((0,4)\) in the xy-plane and let \(B\) be the point \((2 t, 0)\). Let \(L\) be the midpoint of \(A B\) and let the perpendicular bisector of \(A B\) meet the \(y\)-axis \(M\). Let \(N\) be the midpoint of LM. Then locus of \(N\) is

• A

  a circle
• Correct answer
  B

  a parabola
• C

  a straight line
• D

  a hyperbola

Solution

Hint :
Equation of LM \(\begin{aligned} & y-2=\frac{t}{2}(x-t) \\ & \Rightarrow M\left(0, \frac{4-t^2}{2}\right) \end{aligned}\) Midpoint of LM : N(h, k) \(\begin{aligned} & \Rightarrow \quad 2 h=t, 2 k=4-\frac{t^2}{2} \\ & \Rightarrow \quad x^2=2-y \end{aligned}\)

Hence, the correct answer is option: 'B'.