If $A$ is a point $(0, 2)$ and $B$ is a point on the parabola $x^{2} = 4 y$ , then the locus of the midpoint of $A B$ is

x^{2} = 2 (y + 1)

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x^{2} = 4 (y + 1)

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x^{2} = 2 (y - 1)

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x^{2} = (y - 1)

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Solution

The correct option is A $x^{2} = 2 (y + 1)$
$A = (0, 2) B = (h, k) x, y .$
$x^{2} = 4 y \Rightarrow h^{2} = 4 k$
Let $(α, β)$ be M.P of AB
$α = \frac{0 + h}{2}; β = \frac{2 + k}{2}$
$2 α = h; 2 β - 2 = k$
Now $(2 α)^{2} = 4 (2 β - 2)$
$4 α^{2} = 8 β - 8$
$4 x^{2} - 8 y + 8 = 0 \Rightarrow x^{2} = 2 (y - 1) .$

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If A is a point (0,2) and B is a point on the parabola x2=4y, then the locus of the midpoint of AB is

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If $A$ is a point $(0, 2)$ and $B$ is a point on the parabola $x^{2} = 4 y$ , then the locus of the midpoint of $A B$ is