A is (-2, 0) and P is any point on the curve given by `y^(2)=16x`. If Q bisect AP, find the equation of the locus of Q.

Answer

To find the equation of the locus of point Q, which bisects the segment AP where A is (-2, 0) and P is any point on the curve given by \( y^2 = 16x \), we can follow these steps:

### Step 1: Identify the coordinates of point P
The curve \( y^2 = 16x \) is a parabola. We can express the coordinates of point P in terms of a parameter \( t \). For the parabola \( y^2 = 4ax \), where \( a = 4 \), the parametric coordinates are given by:
\[ P(4t^2, 8t) \]

### Step 2: Find the coordinates of point Q
Point Q is the midpoint of segment AP. The coordinates of point A are (-2, 0) and the coordinates of point P are \( (4t^2, 8t) \). The midpoint Q can be calculated using the midpoint formula:
\[ Q\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Substituting the coordinates of A and P:
\[ Q\left( \frac{-2 + 4t^2}{2}, \frac{0 + 8t}{2} \right) = Q\left( \frac{-2 + 4t^2}{2}, 4t \right) \]
This simplifies to:
\[ Q\left( -1 + 2t^2, 4t \right) \]

### Step 3: Express \( t \) in terms of \( k \)
Let \( k = 4t \). Then we can express \( t \) as:
\[ t = \frac{k}{4} \]

### Step 4: Substitute \( t \) back into the equation for Q
Substituting \( t = \frac{k}{4} \) into the x-coordinate of Q:
\[ H = -1 + 2\left(\frac{k}{4}\right)^2 = -1 + 2\left(\frac{k^2}{16}\right) = -1 + \frac{k^2}{8} \]
Thus, we have:
\[ H + 1 = \frac{k^2}{8} \]

### Step 5: Rearranging to find the locus equation
Rearranging gives us:
\[ k^2 = 8(H + 1) \]
Replacing \( H \) with \( x \) and \( k \) with \( y \):
\[ y^2 = 8(x + 1) \]

### Final Equation
The equation of the locus of Q is:
\[ y^2 = 8(x + 1) \]