Question

Let $P$ be a point on $(2, 0)$ and $Q$ be a variable point on $(y - 6)^2 = 2(x - 4)$. Then the locus of the midpoint of $PQ$ is

• A. y2+x+6y+12=0
• B. y2-x+6y+12=0
• C. y2+x-6y+12=0
• D. y2-x-6y+12=0

Answer 1

The Correct Option is D

Given: - Point \( P = (2, 0) \) is fixed - Point \( Q = (x, y) \) lies on the parabola \( (y - 6)^2 = 2(x - 4) \) - We are to find the locus of the midpoint of segment \( PQ \) Step 1: Let midpoint of PQ be M Let the midpoint \( M = (h, k) \) Then: \[ h = \frac{2 + x}{2}, \quad k = \frac{0 + y}{2} \Rightarrow x = 2h - 2, \quad y = 2k \] Step 2: Use the condition that Q lies on the parabola Substitute \( x = 2h - 2 \), \( y = 2k \) into the equation: \[ (y - 6)^2 = 2(x - 4) \Rightarrow (2k - 6)^2 = 2((2h - 2) - 4) \Rightarrow (2k - 6)^2 = 2(2h - 6) \] Step 3: Simplify the equation \[ (2k - 6)^2 = 4h - 12 \Rightarrow 4k^2 - 24k + 36 = 4h - 12 \Rightarrow 4k^2 - 24k - 4h + 48 = 0 \] Divide entire equation by 4: \[ k^2 - 6k - h + 12 = 0 \Rightarrow k^2 + (-h) - 6k + 12 = 0 \Rightarrow k^2 - h - 6k + 12 = 0 \] Now replace back \( h = x \), \( k = y \) to express the locus: \[ y^2 - x - 6y + 12 = 0 \] Final Answer: \[ \boxed{y^2 - x - 6y + 12 = 0} \]