A parabola under a restricted translation

Problem

The vertex $P$ of a parabola $\Gamma$ lies on the line $\ell :y=1+2x$, and the graph meets the $x$-axis at $A=\left(-\frac{1}{2},0\right)$ and $B=\left(\frac{1}{2},0\right)$. Translate the whole parabola so that the new vertex $Q$ is still on $\ell$ and the new graph still passes through $B$. Given that $P$ and $Q$ are distinct, find the distance $PQ$. The full question is on the CEEC 115 GSAT Mathematics A paper.

Where it goes wrong

The item analysis compiles the CEEC report for this question: a correct rate of $20\%$, $47\%$ in the high-scoring group and $3\%$ in the low-scoring group. What is easy to miss is not solving the quadratic. It is the order of the conditions. “Still passes through $B$” gives two positions, and the original position $Q=P$ is one of them. ”$P$ and $Q$ are distinct” is what rules that one out.

Idea

From the two intercepts on the $x$-axis, the original parabola can be written

Its axis of symmetry is $x=0$, and the vertex also lies on $y=1+2x$, so $P=(0,1)$. Substituting gives $a=-4$, that is $f(x)=-4x^2+1$.

Let the translated vertex be $Q=(h,1+2h)$. The new parabola keeps the same shape:

Require that it still pass through the fixed point $B=\left(\frac{1}{2},0\right)$:

So $h=0$ or $h=\frac{3}{2}$. The first makes $Q=P$, which violates that the two points are distinct, so take $h=\frac{3}{2}$. Both vertices move along the direction $(1,2)$, and

The argument uses only the vertex form, translation of a graph, a quadratic equation, and the length of a vector. It does not use a derivative.

What the figure shows

• h=0, Q=(0,1): select the original position. It still passes through $B$, but it violates that $P$ and $Q$ are distinct. The mark is $P=Q$.
• h=3/2, Q=(3/2,4): select the other position. The blue parabola still passes through $B$, and the vertex has moved along $\ell$.
• Compare the segment $PQ$ with the direction of $\ell$, and read the translation vector $(h,2h)$ and the distance $\frac{3\sqrt{5}}{2}$.