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Parabola Vertex Nearest x-axis: p^2 + q^2 = 4 | JEE Main 2026

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

Question
Let the parabola y=x2+px+qy=x^2+px+q pass through the point (1,−1)(1,-1) such that the distance between its vertex and the xx-axis is minimum. Then the value of p2+q2p^2+q^2 is
A22
B44correct
C55
D88
Solution
Step 1: The parabola passes through (1,−1)(1,-1):
−1=1+p+q ⇒ q=−2−p.(1)-1=1+p+q\ \Rightarrow\ q=-2-p.\quad(1)
 Step 2: The vertex's distance from the xx-axis is its ∣y∣|y|-coordinate, ∣−D4a∣\left|\dfrac{-D}{4a}\right| with a=1a=1:
−D4a=−(p2−4q)4=4q−p24.\frac{-D}{4a}=\frac{-(p^2-4q)}{4}=\frac{4q-p^2}{4}.
 Step 3: Substitute q=−2−pq=-2-p from (1):
4(−2−p)−p24=−8−4p−p24=−(p2+4p+8)4=−((p+2)2+4)4.\frac{4(-2-p)-p^2}{4}=\frac{-8-4p-p^2}{4}=\frac{-(p^2+4p+8)}{4}=\frac{-\big((p+2)^2+4\big)}{4}.
 Step 4: Its magnitude (p+2)2+44\dfrac{(p+2)^2+4}{4} is minimum when (p+2)2=0(p+2)^2=0, i.e. p=−2p=-2. Then q=−2−(−2)=0q=-2-(-2)=0. Step 5:
p2+q2=(−2)2+02=4.p^2+q^2=(-2)^2+0^2=4.
 Correct answer: (2)
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