Question

Let \(A(a,0)\) be a point on the \(X\)-axis. The locus of the midpoints of the line segments joining \(A\) and any point \(P\) on the parabola \[ y=16-x^2 \] is

• A. a circle
• B. an ellipse
• C. a pair of lines
• D. a parabola

Hint:

To find the locus of a midpoint, first write the midpoint coordinates in terms of the variable point, then eliminate the parameter to obtain the required equation.

Answer 1

The Correct Option is D

Step 1: Assume a variable point on the parabola. Let \[ P(h,\;16-h^2) \] be any point on the parabola. The fixed point is \[ A(a,0). \]

Step 2: Find the midpoint. Let the midpoint be \[ M(x,y). \] Using the midpoint formula, \[ x=\frac{a+h}{2}, \qquad y=\frac{16-h^2}{2}. \] From the first equation, \[ h=2x-a. \]

Step 3: Obtain the locus. Substituting \[ h=2x-a \] into the equation for \(y\), \[ y = \frac{16-(2x-a)^2}{2}. \] Hence, \[ 2y = 16-(2x-a)^2, \] or \[ (2x-a)^2 = 16-2y. \] This is the equation of a parabola. Therefore, \[ \boxed{\text{The locus is a parabola}.} \] Thus, \[ \boxed{(D)} \] is the correct answer.