AMC10 2015 A

AMC10 2015 A · Q12

AMC10 2015 A · Q12. It mainly tests Quadratic equations, Manipulating equations.

Points $(\sqrt{\pi}, a)$ and $(\sqrt{\pi}, b)$ are distinct points on the graph of $y^2 + x^4 = 2x^2y + 1$. What is $|a-b|$?

点 $(\sqrt{\pi}, a)$ 和 $(\sqrt{\pi}, b)$ 是曲线 $y^2 + x^4 = 2x^2y + 1$ 上的两个不同点。求 $|a-b|$。

(A) 1 1

(B) \frac{\pi}{2} \frac{\pi}{2}

(D) \sqrt{1+\pi} \sqrt{1+\pi}

(E) 1+\sqrt{\pi} 1+\sqrt{\pi}

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): The equation is equivalent to $1=y^{2}-2x^{2}y+x^{4}=(y-x^{2})^{2}$, or $y-x^{2}=\pm 1$. The graph consists of two parabolas, $y=x^{2}+1$ and $y=x^{2}-1$. Thus $a$ and $b$ are $\pi+1$ and $\pi-1$, and their difference is 2. Indeed, the answer would still be 2 if $\sqrt{\pi}$ were replaced by any real number.

答案（C）：该方程等价于 $1=y^{2}-2x^{2}y+x^{4}=(y-x^{2})^{2}$，即 $y-x^{2}=\pm 1$。图像由两条抛物线组成：$y=x^{2}+1$ 与 $y=x^{2}-1$。因此 $a$ 与 $b$ 分别为 $\pi+1$ 和 $\pi-1$，它们的差为 2。事实上，即使把 $\sqrt{\pi}$ 替换为任意实数，答案仍为 2。

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