AMC12 2011 B

AMC12 2011 B · Q14

AMC12 2011 B · Q14. It mainly tests Circle theorems, Trigonometry (basic).

A segment through the focus $F$ of a parabola with vertex $V$ is perpendicular to $\overline{FV}$ and intersects the parabola in points $A$ and $B$. What is $\cos\left(\angle AVB\right)$?

一条经过抛物线焦点 $F$ 的线段与 $\overline{FV}$ 垂直（$V$ 为顶点），并与抛物线交于点 $A$ 和 $B$。求 $\cos\left(\angle AVB\right)$。

(A) -\frac{3\sqrt{5}}{7} -\frac{3\sqrt{5}}{7}

(B) -\frac{2\sqrt{5}}{5} -\frac{2\sqrt{5}}{5}

(D) -\frac{3}{5} -\frac{3}{5}

(E) -\frac{1}{2} -\frac{1}{2}

Answer

Correct choice: (D)

正确答案：（D）

Solution

Name the directrix of the parabola $l$. Define $d(X,k)$ to be the distance between a point $X$ and a line $k$. Now we remember the geometric definition of a parabola: given any line $l$ (called the directrix) and any point $F$ (called the focus), the parabola corresponding to the given directrix and focus is the locus of the points that are equidistant from $F$ and $l$. Therefore $FV=d(V,l)$. Let this distance be $d$. Now note that $d(F,l)=2d$, so $d(A,l)=d(B,l)=2d$. Therefore $AF=BF=2d$. We now use the Pythagorean Theorem on triangle $AFV$; $AV=\sqrt{AF^2+FV^2}=d\sqrt{5}$. Similarly, $BV=d\sqrt{5}$. We now use the Law of Cosines: \[AB^2=AV^2+VB^2-2AV\cdot VB\cos{\angle AVB}\Rightarrow 16d^2=10d^2-10d^2\cos{\angle AVB}\] \[\Rightarrow \cos{\angle AVB}=-\frac{3}{5}\] This shows that the answer is $\boxed{\textbf{(D)}}$.

设抛物线的准线为 $l$。定义 $d(X,k)$ 为点 $X$ 到直线 $k$ 的距离。回忆抛物线的几何定义：给定一条直线 $l$（准线）和一点 $F$（焦点），对应的抛物线是到 $F$ 与到 $l$ 距离相等的点的轨迹。因此 $FV=d(V,l)$。设该距离为 $d$。注意到 $d(F,l)=2d$，所以 $d(A,l)=d(B,l)=2d$。因此 $AF=BF=2d$。在三角形 $AFV$ 中用勾股定理：$AV=\sqrt{AF^2+FV^2}=d\sqrt{5}$。同理 $BV=d\sqrt{5}$。再用余弦定理： \[AB^2=AV^2+VB^2-2AV\cdot VB\cos{\angle AVB}\Rightarrow 16d^2=10d^2-10d^2\cos{\angle AVB}\] \[\Rightarrow \cos{\angle AVB}=-\frac{3}{5}\] 因此答案是 $\boxed{\textbf{(D)}}$。

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