Question

The turning point (vertex) of the parabola $y=x^2-2 a x+1$ 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 $y=x^2-2 a x+1$
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

$\left[-\frac{b}{2 a}, f\left(-\frac{b}{2 a}\right)\right]$
$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 $( C )$