The minimum distance of the point (0, 5) from the parabola x^2 = 4y is

The correct answer is Option (2) → 4

Let (x, y) be a point on the parabola, so \(x^{2} = 4y\) and \(y \ge 0\). The square of its distance from (0, 5) is

    \( D = x^{2} + (y - 5)^{2} \)

    \( = 4y + (y - 5)^{2} \)

Minimising the square of the distance gives the same point:

    \( D'(y) = 4 + 2(y - 5) = 0 \)

    \( y = 3 \)

Since \(D''(y) = 2 \gt 0\), this is the minimum. The points are \((\pm 2\sqrt{3}, 3)\), and

    \( D(3) = 12 + 4 = 16 \)

So the minimum distance is \(\sqrt{16} = 4\).

Why the other options are incorrect

• Option (1) 5 — Incorrect. This is the distance to the vertex (0, 0), which is not the nearest point.

• Option (3) \(2\sqrt{3}\) — Incorrect. This is the x-coordinate of the nearest point, not the distance.

• Option (4) 3 — Incorrect. This is the y-coordinate of the nearest point, not the distance.