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Minimum of SQ+PQ for Point Inside x²+4y=0 | JEE Parabola

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

Question
Consider the parabola x2+4y=0x^2 + 4y = 0. Let P(a,b)P(a, b) be a fixed point inside the parabola and let SS be the focus. Then the minimum value of SQ+PQSQ + PQ, as the point QQ moves on the parabola, is:
A∣1−a∣|1 - a|
B∣ab∣+1|ab| + 1
Ca2+b2\sqrt{a^2 + b^2}
D1−b1 - bcorrect
Solution
Step 1: For x2=−4yx^2 = -4y the focus is S(0,−1)S(0, -1) and the directrix y=1y = 1. If NN is the foot of perpendicular from QQ to the directrix, then SQ=QNSQ = QN, so SQ+PQ=QN+PQSQ + PQ = QN + PQ. Step 2: This is least when P,Q,NP, Q, N are collinear (perpendicular to the directrix), giving the distance from PP to the directrix:
min⁡(SQ+PQ)=1−b.\min(SQ + PQ) = 1 - b.
 Correct answer: (4)
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