AMC12 2022 B

AMC12 2022 B · Q8

AMC12 2022 B · Q8. It mainly tests Graphs (coordinate plane), Geometry misc.

What is the graph of $y^4+1=x^4+2y^2$ in the coordinate plane?

坐标平面中$y^4+1=x^4+2y^2$的图像是什么？

(A) \text{two intersecting parabolas} \text{two intersecting parabolas}

(B) \text{two nonintersecting parabolas} \text{two nonintersecting parabolas}

(D) \text{a circle and a hyperbola} \text{a circle and a hyperbola}

(E) \text{a circle and two parabolas} \text{a circle and two parabolas}

Answer

Correct choice: (D)

正确答案：（D）

Solution

Since the equation has even powers of $x$ and $y$, let $y'=y^2$ and $x' = x^2$. Then $y'^2 + 1 = x'^2 + 2y'$. Rearranging gives $y'^2 - 2y' + 1 = x'^2$, or $(y'-1)^2=x'^2$. There are two cases: $y' \leq 1$ or $y' > 1$. If $y' \leq 1$, taking the square root of both sides gives $1 - y' = x'$, and rearranging gives $x' + y' = 1$. Substituting back in $x'=x^2$ and $y'=y^2$ gives us $x^2+y^2=1$, the equation for a circle. Similarly, if $y' > 1$, we take the square root of both sides to get $y' - 1 = x'$, or $y' - x' = 1$, which is equivalent to $y^2 - x^2 = 1$, a hyperbola. Hence, our answer is $\boxed{\textbf{(D)}\ \text{a circle and a hyperbola}}$. (Solutions 1 and 2 are in essence the same; Solution 1 lets $(x',y')=\left(x^2,y^2\right)$ for convenience, but the two solutions are otherwise identical.)

由于方程中$x$和$y$的幂次都是偶数，设$y'=y^2$，$x' = x^2$。则$y'^2 + 1 = x'^2 + 2y'$。重排得$y'^2 - 2y' + 1 = x'^2$，即$(y'-1)^2=x'^2$。有两个情况：$y' \leq 1$或$y' > 1$。若$y' \leq 1$，两边取平方根得$1 - y' = x'$，重排得$x' + y' = 1$。代回$x'=x^2$，$y'=y^2$得$x^2+y^2=1$，这是一个圆的方程。类似地，若$y' > 1$，取平方根得$y' - 1 = x'$，即$y' - x' = 1$，等价于$y^2 - x^2 = 1$，这是一个双曲线。故答案是$\boxed{\textbf{(D)}\ \text{a circle and a hyperbola}}$。（解法1和2本质相同；解法1为了方便设$(x',y')=\left(x^2,y^2\right)$，但两解否则相同。）

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