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Maths Question 12 – JEE-MAIN 2026

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Let the parabola y=x2+px+q passing through the point (1,−1) be such that the distance between its vertex and the x-axis is minimum. Then the value of p2+q2 is:

Recall the formula for the coordinates of the vertex of a parabola y=ax2+bx+c.

● Video Walkthrough
Step 1: Determine Vertex Coordinates and Constraint✦ Active

The given parabola is y=x2+px+q. The coordinates of its vertex are V(−p2,q−p24). Since the parabola passes through the point (1,−1), we substitute these coordinates into the equation:

−1=(1)2+p(1)+q⟹−1=1+p+q⟹p+q=−2
Step 2: Express Vertex y-coordinate in terms of one variable○ Expand

From the constraint p+q=−2, we can express q as q=−2−p. Now, substitute this into the y-coordinate of the vertex, yv=q−p24:

yv=(−2−p)−p24=−p24−p−2

This can be rewritten as yv=−14(p2+4p+8).

Step 3: Minimize Distance and Calculate Result○ Expand

The distance between the vertex and the x-axis is |yv|. We need to minimize this distance. The expression for yv is a quadratic in p that opens downwards. Its maximum value occurs at p=−42(1)=−2. At this value of p:

yv=−14((−2)2+4(−2)+8)=−14(4−8+8)=−14(4)=−1

Since yv=−1 is the maximum value of yv, and yv is always less than or equal to −1, the minimum value of |yv| is |−1|=1. This occurs when p=−2. Now, use p=−2 in the relation p+q=−2 to find q:

−2+q=−2⟹q=0

Finally, calculate p2+q2:

p2+q2=(−2)2+(0)2=4+0=4
💡 Teacher's Secret Hint

Remember that for a parabola y=ax2+bx+c with a>0, the vertex is the minimum point. Thus, yv is the minimum value of y. To minimize |yv|, we need yv to be as close to zero as possible.

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