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The parabola $y = 2 - x^{2}$ and the line $y = 2 x + 4$ are drawn in the plane.

The points $P$ and $Q$ lie on the parabola and the line respectively, such that $P Q$ is a vertical line. The length of $P Q$ is $d$ .

Find the smallest possible value of $d$ .

Hint

Since $Q$ lies above $P$ , the distance $d$ is given by the $y$ coordinate of $Q$ minus the $y$ coordinate of $P$

Line and parabola

Hint

d = (2 x + 4) - (2 - x^{2})

Solution

We have

\begin{aligned} d & = y_{Q} - y_{P} \\ = (2 x + 4) - (2 - x^{2}) \\ = x^{2} + 2 x + 2 \end{aligned}

First we note that $Δ = 4 - 4 (1) (2) < 0$ so we never have $d = 0$ , which is to say that the parabola and line never intersect.

Let's find the minimum value of $d$ by considering the complete square form

\begin{aligned} d & = x^{2} + 2 x + 2 \\ = (x + 1)^{2} - 1 + 2 \\ = (x + 1)^{2} + 1 \end{aligned}

So when $x = - 1$ , $d$ achieves its minimum value of $1$ .