Question
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 :
Options
Solution
Sol. [Key Concept: Vertex ordinate $= - D/4a$; minimize over the constraint $p + q = - 2$, then $p^{2} + q^{2}$.]
parabola passes through $(1, - 1)$ so
$${- 1 = 1 + p + q }{p + q = - 2}$$
$$\begin{array}{r} q = - 2 - p\#(1) \end{array}$$
Distance from x -axis
$$\begin{matrix} & \frac{- D}{4a} = \frac{- \left( p^{2} - 4q \right)}{4(1)} = \frac{4q - p^{2}}{4} \\ & \ = \frac{4( - 2 - p) - p^{2}}{4}\ \text{~}\text{from (1)}\text{~} \\ & \ = \frac{- 8 - 4p - p^{2}}{4} = \frac{- \left( p^{2} + 4p + 8 \right)}{4} = \frac{- \left( (p + 2)^{2} + 4 \right)}{4} \end{matrix}$$
Minimum at $p = - 2 \Rightarrow q = 0$
$$p^{2} + q^{2} = 4 + 0 = 4 $$
Create a free account to view solution
View Solution Free