Let the parabola y = x^2 + 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 p^2 + q^2 is:

Step-by-step solution

Given the equation of the parabola y = x^2 + px + q . Since it passes through (1, -1) , we have: -1 = 1 + p + q q = -p - 2 The vertex of the parabola y = x^2 + px + q occurs at x = - p 2 . The y -coordinate of the vertex is: y_v = (- p 2 )^2 + p (- p 2 ) + q = - p^2 4 + q The distance between the vertex and the x -axis is |y_v| . Substituting q = -p - 2 into the expression for y_v : |y_v| = |- p^2 4 - p - 2 | = 1 4 |p^2 + 4p + 8| = 1 4 |(p+2)^2 + 4| To minimize this distance, the term (p+2)^2 must be minimum. Since