2015 AMC 10A Problem 12

Group the terms involving $y$ and $x^2$ together:
$$y^2 - 2x^2 y + x^4 = 1 \quad\Longrightarrow\quad (y - x^2)^2 = 1.$$
So every point of the graph satisfies $y - x^2 = \pm 1$, i.e. the graph is the union of the two parabolas $y = x^2 + 1$ and $y = x^2 - 1$.

For a fixed $x$ (here $x = \sqrt{\pi}$, so $x^2 = \pi$) the two points on the graph have $y$-coordinates $\pi + 1$ and $\pi - 1$. Since the points are distinct, $\{a, b\} = \{\pi + 1, \pi - 1\}$ and
$$|a - b| = (\pi + 1) - (\pi - 1) = 2.$$

The answer is $\boxed{\textbf{(C)}\ 2}$.

The expression $y^2 - 2x^2y + x^4$ is $(y - x^2)^2$, a perfect square in the "variables" $y$ and $x^2$. Recognizing that structure turns a scary quartic into two familiar parabolas one unit apart. The specific $\sqrt{\pi}$ is a decoy: the vertical gap is $2$ at every $x$.

Substitute $x^2 = \pi$ directly: $y^2 - 2\pi y + (\pi^2 - 1) = 0$. By Vieta, $a + b = 2\pi$ and $ab = \pi^2 - 1$, so $(a - b)^2 = (a+b)^2 - 4ab = 4\pi^2 - 4\pi^2 + 4 = 4$ and $|a - b| = 2$.

• Plugging in $x = \sqrt{\pi}$ and trying to solve $y^2 - 2\pi y + \pi^2 = 1$ by the quadratic formula, where the $\pi$'s invite arithmetic errors.
• Taking only the positive square root ($y = x^2 + 1$) and then being unable to find a second point.