`P(x , y)` is a variable point on the parabola `y^2=4a x` and `Q(x+c ,y+c)` is another variable point, where `c` is a constant. The locus of the midpoint of `P Q` is a/n parabola (b) hyperbola hyperbola (d) circle

Options

A. parabola
B. ellipse
C. hyperbola
D. circle

Explanation

To find the locus of the midpoint of points \( P(x, y) \) and \( Q(x+c, y+c) \) where \( P \) lies on the parabola \( y^2 = 4ax \), we will follow these steps:

### Step 1: Identify the midpoint coordinates
The midpoint \( M \) of points \( P \) and \( Q \) can be calculated using the midpoint formula:
\[ M\left(h, k\right) = \left(\frac{x + (x+c)}{2}, \frac{y + (y+c)}{2}\right) \]
This simplifies to:
\[ h = \frac{2x + c}{2} \quad \text{and} \quad k = \frac{2y + c}{2} \]

### Step 2: Express \( x \) and \( y \) in terms of \( h \) and \( k \)
From the equations for \( h \) and \( k \):
\[ 2h = 2x + c \implies 2x = 2h - c \implies x = h - \frac{c}{2} \]
\[ 2k = 2y + c \implies 2y = 2k - c \implies y = k - \frac{c}{2} \]

### Step 3: Substitute \( x \) and \( y \) into the parabola equation
Since point \( P \) lies on the parabola \( y^2 = 4ax \), we substitute \( x \) and \( y \) in terms of \( h \) and \( k \) into this equation:
\[ \left(k - \frac{c}{2}\right)^2 = 4a\left(h - \frac{c}{2}\right) \]

### Step 4: Expand and rearrange the equation
Expanding the left side:
\[ \left(k - \frac{c}{2}\right)^2 = k^2 - ck + \frac{c^2}{4} \]
And the right side:
\[ 4a\left(h - \frac{c}{2}\right) = 4ah - 2ac \]
Setting the two sides equal gives:
\[ k^2 - ck + \frac{c^2}{4} = 4ah - 2ac \]

### Step 5: Rearranging to find the locus
Rearranging the equation, we have:
\[ k^2 - ck - 4ah + 2ac + \frac{c^2}{4} = 0 \]
This is a quadratic equation in \( k \). The general form of a parabola is \( (k - k_0)^2 = 4p(h - h_0) \).

### Conclusion
The locus of the midpoint \( M \) is a parabola.