PART I · SHORT RESPONSE第一部分 · 短答题SAT-style MCQ + ON / AB short answer · 18 marksSAT 风格选择题 + 安 / 阿省考短答 · 共 18 分
Section A · Short ResponseA 部分 · 短答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter; show enough work in the margin that a marker could verify. State the standard form $(x-h)^2 + (y-k)^2 = r^2$ or its conic analogue explicitly before reading off centre / radius / foci. No calculator on Q1–Q4; calculator permitted on Q5.本节包含选择题与短答题。选择题请圈出字母答案，并在空白处写下足够过程供阅卷人核验。在读出中心 / 半径 / 焦点前，需先明确写出标准方程 $(x-h)^2 + (y-k)^2 = r^2$ 或相应圆锥曲线的标准形式。Q1–Q4 不可使用计算器；Q5 可用计算器。
Q1EASY易🇺🇸 US美SAT-style MCQSAT 风格选择题§1 Circle Standard Form圆的标准式 · HSG-GPE.A.1[3 marks][3 分]
In the $xy$-plane, the equation $(x - 2)^{2} + (y + 5)^{2} = 49$ describes a circle. What are the centre and radius of this circle?在 $xy$ 平面内，方程 $(x - 2)^{2} + (y + 5)^{2} = 49$ 表示一个圆。该圆的圆心与半径是？
A parabola in the $xy$-plane has focus $(0, 3)$ and directrix $y = -3$. Which of the following is its equation?$xy$ 平面内一条抛物线的焦点为 $(0, 3)$，准线为 $y = -3$。其方程是下列哪一项？
An ellipse centred at the origin has equation $\dfrac{x^{2}}{25} + \dfrac{y^{2}}{9} = 1$. What is its eccentricity $e$?中心位于原点的椭圆方程为 $\dfrac{x^{2}}{25} + \dfrac{y^{2}}{9} = 1$。其离心率 $e$ 是多少？
The points $A(-1, 2)$ and $B(7, 8)$ are the endpoints of a diameter of a circle in the $xy$-plane.$xy$ 平面内，点 $A(-1, 2)$ 与 $B(7, 8)$ 是某圆一条直径的两端点。
(a)Determine the centre of the circle (midpoint of $AB$).求该圆的圆心（即 $AB$ 的中点）。[1]
(b)Determine the radius of the circle.求该圆的半径。[2]
(c)State the equation of the circle in standard form $(x - h)^{2} + (y - k)^{2} = r^{2}$.写出该圆的标准式方程 $(x - h)^{2} + (y - k)^{2} = r^{2}$。[1]
Q5MEDIUM中🇨🇦 AB阿AB Provincial-style阿尔伯塔省考风格§2 Parabola, Vertex Form抛物线顶点式 · AB Math 20-1 RF GO 3[5 marks][5 分]
A parabola has vertex $(2, -3)$ and opens upward with $a = \dfrac{1}{4}$ in the form $y = a(x - h)^{2} + k$. Use the focus-directrix relation $4p = \dfrac{1}{a}$ (with focus $(h, k + p)$ and directrix $y = k - p$) to answer the following.某抛物线顶点为 $(2, -3)$，开口向上，形式为 $y = a(x - h)^{2} + k$ 且 $a = \dfrac{1}{4}$。利用焦点—准线关系 $4p = \dfrac{1}{a}$（焦点为 $(h, k + p)$，准线为 $y = k - p$）回答以下问题。
(a)Write the equation of the parabola in vertex form.写出该抛物线的顶点式方程。[1]
(b)Determine the focal length $p$.求焦距 $p$。[1]
(c)State the coordinates of the focus and the equation of the directrix.写出焦点的坐标与准线的方程。[2]
(d)State the equation of the axis of symmetry.写出对称轴的方程。[1]
PART II · EXTENDED RESPONSE第二部分 · 拓展题AP-feeder FRQ + honors · 35 marksAP 衔接简答题 + 荣誉级 · 共 35 分
Section B · Extended ResponseB 部分 · 拓展题
Show every algebraic step. State the standard form of the conic (circle, ellipse, hyperbola, or parabola) before reading off centre, vertices, foci, or asymptotes. Use $a^{2}$, $b^{2}$, $c^{2}$ consistently and cite the focal relation ($c^{2} = a^{2} - b^{2}$ for ellipses, $c^{2} = a^{2} + b^{2}$ for hyperbolas) explicitly. No calculator on Q6–Q9 unless noted.写出每一步代数过程。在读出中心、顶点、焦点或渐近线前，先明确写出该圆锥曲线（圆、椭圆、双曲线或抛物线）的标准式。$a^{2}$、$b^{2}$、$c^{2}$ 的使用要前后一致，并显式引用焦点关系（椭圆 $c^{2} = a^{2} - b^{2}$，双曲线 $c^{2} = a^{2} + b^{2}$）。Q6–Q9 除非另注明，否则不可使用计算器。
An ellipse centred at the origin has its major axis along the $x$-axis, semi-major axis length $a = 5$, and semi-minor axis length $b = 4$.中心位于原点的椭圆，长轴沿 $x$ 轴，半长轴长度 $a = 5$，半短轴长度 $b = 4$。
(a)Write the equation of the ellipse in standard form $\dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} = 1$.写出椭圆的标准式方程 $\dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} = 1$。[1]
(b)State the coordinates of the four vertices (endpoints of the major and minor axes).写出四个顶点（长轴顶点与短轴端点）的坐标。[2]
(c)Use the focal relation $c^{2} = a^{2} - b^{2}$ to find $c$, then state the coordinates of the two foci.利用焦点关系 $c^{2} = a^{2} - b^{2}$ 求出 $c$，再写出两个焦点的坐标。[3]
(d)Compute the eccentricity $e = c/a$. Briefly classify the ellipse as "nearly circular" or "elongated" based on $e$.计算离心率 $e = c/a$，并据此简要判断该椭圆是"接近圆形"还是"较为扁长"。[2]
Consider the hyperbola $\dfrac{x^{2}}{9} - \dfrac{y^{2}}{16} = 1$. Note: hyperbolas are not in the standard BC PC 11 / PC 12 syllabus; this is an honors-extension question for students continuing to first-year university analytic geometry.考虑双曲线 $\dfrac{x^{2}}{9} - \dfrac{y^{2}}{16} = 1$。说明：双曲线不在 BC PC 11 / PC 12 标准大纲中；本题属荣誉级拓展题，面向衔接大一解析几何的学生。
(a)State $a$, $b$, and the orientation of the transverse axis (horizontal or vertical).写出 $a$、$b$ 以及实轴方向（水平或竖直）。[2]
(b)State the coordinates of the two vertices (endpoints of the transverse axis).写出两个顶点（实轴端点）的坐标。[1]
(c)Use the hyperbola focal relation $c^{2} = a^{2} + b^{2}$ to find $c$, then state the coordinates of the two foci.利用双曲线的焦点关系 $c^{2} = a^{2} + b^{2}$ 求出 $c$，再写出两个焦点的坐标。[3]
(d)State the equations of the two asymptotes (lines $y = \pm (b/a) x$ for a horizontal transverse axis).写出两条渐近线的方程（实轴水平时为 $y = \pm (b/a) x$）。[2]
(e)Briefly explain in one sentence why the focal-relation sign differs between ellipse ($c^{2} = a^{2} - b^{2}$) and hyperbola ($c^{2} = a^{2} + b^{2}$).用一句话简述椭圆（$c^{2} = a^{2} - b^{2}$）与双曲线（$c^{2} = a^{2} + b^{2}$）焦点关系的符号为何不同。[1]
(d)For the equation in part (a), explain in one sentence why the test $A = C$ would (or would not) further reduce the ellipse to a circle.就 (a) 中的方程，用一句话说明 $A = C$ 检验为何能（或不能）进一步把该椭圆退化为圆。[2]
(a)Complete the square in $x$ and in $y$ separately to convert the equation into the standard form $(x - h)^{2} + (y - k)^{2} = r^{2}$. Show each add-and-subtract step explicitly.分别对 $x$ 与 $y$ 配方，将方程化为标准式 $(x - h)^{2} + (y - k)^{2} = r^{2}$。每一步加减都要明确写出。[4]
(b)State the centre $(h, k)$ and the radius $r$.写出圆心 $(h, k)$ 与半径 $r$。[2]
(c)Determine whether the point $(7, -4)$ lies inside, on, or outside the circle. Justify using the standard form.判断点 $(7, -4)$ 在圆内、圆上还是圆外，并用标准式说明理由。[2]
(d)Briefly describe the geometric transformation that takes the unit circle $x^{2} + y^{2} = 1$ to the circle in part (a).简要描述把单位圆 $x^{2} + y^{2} = 1$ 变为 (a) 中之圆的几何变换。[1]
PART III · MODELING / APPLIED第三部分 · 建模与应用Universal · 28 marks通用 · 共 28 分
Section C · Modeling and ApplicationsC 部分 · 建模与应用
Name your variables (with units) before writing equations. Place the conic in a convenient coordinate system (state your axis choice), set up the standard form, solve, and conclude with a one-sentence answer in context. Calculator permitted throughout Part III.列方程前先命名变量并标注单位。把圆锥曲线放入方便的坐标系（写明坐标轴的选取），列出标准式，求解，最后用一句话结合情境作答。第三部分全程可使用计算器。
A parabolic satellite dish has a circular rim of diameter $1.2$ m and a depth of $0.15$ m at its centre. Place the vertex of the parabola at the origin with the axis of symmetry along the positive $y$-axis, so the parabola has the form $y = a x^{2}$. The dish's receiver is mounted at the focus of the parabola.一台抛物面卫星碟形天线圆形边缘直径为 $1.2$ 米，中心深度为 $0.15$ 米。把抛物线的顶点放在原点，对称轴沿 $y$ 轴正方向，故抛物线形式为 $y = a x^{2}$。该碟的接收器安装在抛物线的焦点处。
(a)State the coordinates of the rim's edge point in this coordinate system (the point on the parabola directly above the rim's edge).在该坐标系中写出碟边缘点的坐标（即抛物线上正位于边缘正上方的点）。[1]
(b)Substitute this point into $y = a x^{2}$ and solve for $a$.将该点代入 $y = a x^{2}$ 解出 $a$。[2]
(c)Use $4p = 1/a$ to determine the focal length $p$, in metres (rounded to the nearest centimetre).利用 $4p = 1/a$ 求出焦距 $p$（以米为单位，精确到厘米）。[3]
(d)State the height (above the vertex of the dish) at which the receiver should be mounted, and explain in one sentence why parallel incoming radio waves are concentrated at this point.写出接收器应安装在距碟顶多高（顶点上方），并用一句话解释为什么平行入射的无线电波会汇聚到该点。[3]
A whisper gallery is built in the shape of half an ellipse, with its major axis horizontal along the floor. The gallery is $24$ m long (the full length of the major axis) and $7$ m tall at its centre (the semi-minor axis). Two visitors stand at the foci of the ellipse so that any whisper from one focus reflects off the ceiling and arrives clearly at the other. Ellipse foci / Kepler orbits are not in standard AB Math 20-1 / 30-1; this is honors-extension content.某回音廊建成半椭圆形，长轴沿地面水平。回廊长 $24$ 米（即长轴全长），中心处高 $7$ 米（即半短轴）。两位访客分别站在椭圆的两焦点上，使一方在焦点处的低语经天花板反射后能清晰传至另一焦点。椭圆焦点 / 开普勒轨道不在阿尔伯塔 Math 20-1 / 30-1 标准大纲内，属荣誉级拓展内容。
(a)Place the centre of the ellipse at the origin with the major axis along the $x$-axis. State $a$ and $b$ (in metres).把椭圆中心放在原点，长轴沿 $x$ 轴。写出 $a$ 与 $b$（单位：米）。[1]
(b)Use the focal relation $c^{2} = a^{2} - b^{2}$ to compute $c$, the distance from the centre to each focus. Round to the nearest tenth of a metre.利用焦点关系 $c^{2} = a^{2} - b^{2}$ 求出 $c$，即中心到每个焦点的距离。结果保留到 $0.1$ 米。[3]
(c)State the coordinates of the two foci (where the visitors should stand) and the distance between them.写出两个焦点（即访客站立位置）的坐标以及两焦点之间的距离。[2]
(d)Compute the eccentricity $e = c/a$ of the elliptical cross-section. State, in one sentence, what range of $e$ values would make the gallery "nearly circular" (poor whisper effect) versus "elongated" (strong whisper effect).计算椭圆横截面的离心率 $e = c/a$。用一句话说明：$e$ 取何范围会使回廊"接近圆形"（回音效果差）、何范围会"较为扁长"（回音效果强）。[3]
In a LORAN (Long-Range Navigation) system, a ship measures the time difference between two synchronized radio pulses from two transmitter stations. The set of points whose difference of distances to the two stations is a constant traces out one branch of a hyperbola. Two LORAN stations are located at $F_{1}(-200, 0)$ and $F_{2}(200, 0)$ (coordinates in kilometres). A ship records that its distance to $F_{1}$ is $240$ km greater than its distance to $F_{2}$. Hyperbolas / LORAN are not in standard BC PC 11 / PC 12; this is honors-extension content.在 LORAN（远程导航）系统中，船只测量来自两座同步无线电发射台脉冲的时间差。到两台距离之差为常数的点集恰为双曲线的一支。两座 LORAN 站分别位于 $F_{1}(-200, 0)$ 与 $F_{2}(200, 0)$（坐标单位：公里）。某船记录到自己到 $F_{1}$ 的距离比到 $F_{2}$ 的距离多 $240$ 公里。双曲线 / LORAN 不在卑诗 PC 11 / PC 12 标准大纲内，属荣誉级拓展内容。
(a)State the focal distance $c$ (in kilometres) and the constant difference $2a$, then deduce $a$.写出焦距 $c$（单位：公里）与常数距离差 $2a$，再推出 $a$。[2]
(b)Use the hyperbola focal relation $c^{2} = a^{2} + b^{2}$ to compute $b^{2}$.利用双曲线焦点关系 $c^{2} = a^{2} + b^{2}$ 求出 $b^{2}$。[2]
(c)State the equation of the hyperbola (centred at the origin, transverse axis along the $x$-axis) in standard form $\dfrac{x^{2}}{a^{2}} - \dfrac{y^{2}}{b^{2}} = 1$.写出该双曲线的标准方程（中心在原点、实轴沿 $x$ 轴）：$\dfrac{x^{2}}{a^{2}} - \dfrac{y^{2}}{b^{2}} = 1$。[1]
(d)Determine on which branch (left, $x \le -a$, or right, $x \ge a$) the ship's location lies, given that the distance to $F_{1}$ exceeds the distance to $F_{2}$. Justify briefly.已知该船到 $F_{1}$ 的距离大于到 $F_{2}$ 的距离，判断船位于哪一支（左支 $x \le -a$ 还是右支 $x \ge a$），并简要说明理由。[2]
(e)If a second pair of LORAN stations independently locates the ship on another hyperbola, explain in one sentence how the ship's exact position is determined.若第二对 LORAN 站独立确定该船位于另一条双曲线上，用一句话说明如何由此确定船只的精确位置。[1]
(f)State the equations of the two asymptotes of the LORAN hyperbola from part (c). Round slopes to three decimal places.写出 (c) 中 LORAN 双曲线的两条渐近线方程。斜率保留三位小数。[2]
🇺🇸 US Common Core美国共同核心HSG-GPE.A.1(circle)（圆） · HSG-GPE.A.2(parabola, focus-directrix)（抛物线，焦点—准线） · HSG-GPE.A.3 (+)(ellipse / hyperbola, foci sum / difference)（椭圆 / 双曲线，焦点距离之和 / 差）
🇨🇦 Ontario安大略MPM2D analytic geometry of the circle only · MCR3U / MHF4U treat parabola only as a quadratic; ellipse + hyperbola are out-of-scope (honors / extension)MPM2D 仅涉及圆的解析几何 · MCR3U / MHF4U 仅把抛物线当作二次函数处理；椭圆与双曲线超出大纲（荣誉级 / 拓展内容）
🇨🇦 British Columbia不列颠哥伦比亚PC 11 covers quadratic = parabola only · PC 12 does not formalise the four-conic taxonomy; ellipse + hyperbola + discriminant are honors / extension contentPC 11 仅涉及二次函数即抛物线 · PC 12 未正式建立四类圆锥曲线分类；椭圆、双曲线与判别式属荣誉级 / 拓展内容
🇨🇦 Alberta阿尔伯塔Math 20-1 RF (Relations & Functions) GO 3 covers parabolas as quadratic graphs; ellipse + hyperbola not in Math 20-1 / 30-1 (honors / extension)Math 20-1 RF（关系与函数）GO 3 将抛物线作为二次函数图像处理；椭圆与双曲线不在 Math 20-1 / 30-1 中（荣誉级 / 拓展内容）
Full 4-column Syllabus Map lives in ../Study Guides/Unit_12_Conic_Sections.html. Honors flag attached to ON / BC / AB rows on ellipse, hyperbola, and discriminant items, where the topic sits outside the standard provincial syllabus.完整的四列大纲对照表见 ../Study Guides/Unit_12_Conic_Sections.html。安 / 卑 / 阿三省在椭圆、双曲线、判别式题目上均标注荣誉级标签，因为这些内容不在各省标准大纲中。