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Locus of Midpoint of Focal Segment of y^2 = 8x | JEE Advanced Parabola

JEE Maths question with a full step-by-step solution.

Question
A point moves on the parabola y2=8xy^{2} = 8x and is joined to the focus of that parabola. Find the equation of the locus of the mid point of this segment.
Ay2=4x−1y^{2} = 4x - 1
By2=4x−4y^{2} = 4x - 4correct
Cx2=4y−1x^{2} = 4y - 1
Dx2+y2=1x^{2} + y^{2} = 1
Solution
Step 1: y2=8xy^{2}=8x is y2=4axy^{2}=4ax with a=2a=2, so the focus is S(2,0)S(2,0). A general point of the parabola is
P(2t2, 4t)P\left(2t^{2},\,4t\right)
 Step 2: The mid point (h,k)(h,k) of SPSP:
h=2+2t22=1+t2,k=0+4t2=2th = \frac{2 + 2t^{2}}{2} = 1 + t^{2}, \qquad k = \frac{0 + 4t}{2} = 2t
 Step 3: From the second relation t=k2t = \dfrac{k}{2}, so
h=1+k24 ⟹ 4h−4=k2h = 1 + \frac{k^{2}}{4} \ \Longrightarrow\ 4h - 4 = k^{2}
 Step 4: Replacing (h,k)(h,k) by (x,y)(x,y),
y2=4x−4y^{2} = 4x - 4
 Answer: (2).
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