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 :
- 2
- 4
- 5
- 8
Answer
(2) 4
Solution
(1, -1) lies on the parabola: 1+p+q=-1 q=-p-2
Vertex: (- p2, q- p24 )
Ordinate of vertex =-p-2- p24=- p2+4p+84=- (p+2)2+44
Distance of vertex from the x-axis = (p+2)2+44 1
Minimum when p=-2 q=0
p2+q2=4+0=4
Answer: (2)