Question

The turning point (vertex) of the parabola $y=x^2-2ax+1$ is closest to the origin when

• A. $a=0$
• B. $a=\pm 1$
• C. $a=\pm \frac{1}{\sqrt{2}}$
• D. $a=\pm \frac{1}{\sqrt{2}}$ or $a=0$

Answer 1

Answer: (C)

As $y=x^2-2ax+1$
then the parabola's tuming point is at the vertex $\left(a,1-a^2\right)$
The distance $D$ from the origin to this point is

$\left[-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right]$
$D^2=a^2+{\left(1-a^2\right)}^2$
$=a^4-a^2+1$
$={\left(a^2-\frac{1}{2}\right)}^2+\frac{3}{4}$
so $D^2$, and hence $D$, is at a minimum when $a^2=\frac{1}{2}$. The Answer is $(C)$