High School Math

Conic Sections圆锥曲线

Slice a double cone with a plane and you get one of four curves: a circle, an ellipse, a parabola, or a hyperbola. The same four curves admit purely planar definitions in terms of distances to one or two fixed points (foci) and sometimes a fixed line (directrix). This guide walks the algebra of all four: standard-form equations, how to read centre / vertices / foci / asymptotes off the equation, how to identify a curve from the general second-degree form $Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0$ via the discriminant $B^{2} - 4AC$, how to shift any conic to a translated centre, and where the curves show up in the world, parabolic dishes, planetary orbits (Kepler), whisper galleries, and hyperbolic navigation systems.用一个平面去切双锥面,所得截线只可能是四种曲线之一:圆、椭圆、抛物线、双曲线(circleellipseparabolahyperbola)。同样这四条曲线也有纯平面定义,以到一或两个定点(焦点 focus)以及有时一条定直线(准线 directrix)的距离来刻画。本指南把四条曲线的代数走一遍:标准方程;如何从方程读出中心 / 顶点 / 焦点 / 渐近线;如何用判别式 $B^{2} - 4AC$ 从一般二次式 $Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0$ 辨识曲线类型;如何把任一圆锥曲线平移到新中心;以及这些曲线在现实中的位置,抛物面碟、行星轨道(开普勒)、回音廊、双曲线导航系统。

7 sections7 节内容 US Common Core · ON · BC · ABUS 共同核心 · ON · BC · AB Honors block on Ellipses / Hyperbolas / Discriminant椭圆 / 双曲线 / 判别式为荣誉级

How to use this guide如何使用本指南

Conic sections is one of the curricula-divergent units in this series. In the US, all four curves , circle, parabola, ellipse, hyperbola , appear in CCSSM Geometry under HSG-GPE.A.1, .A.2, and .A.3 (with .A.3 marked (+) honors), and the full set is required in AP-feeder Pre-Calc. In Canada the published Grades 10-12 curricula for Ontario, BC, and Alberta treat only the circle and the parabola, and only as algebraic objects (circle equation, vertex-form quadratic), not as focus / directrix loci. Ellipses, hyperbolas, eccentricity, and the discriminant-of-the-general-quadratic test (§5) are not in any of the three Canadian curricula at core level. The eight-row table tells you which sections are on your syllabus right now.圆锥曲线是本系列中"大纲分歧最大"的单元之一。美国 CCSSM 几何课在 HSG-GPE.A.1、.A.2、.A.3 下覆盖四种曲线(圆、抛物线、椭圆、双曲线;其中 .A.3 标 (+) 荣誉级),并在 AP 衔接 Pre-Calc 中要求完整掌握。加拿大三省(安、BC、阿)已公布的 10-12 年级课纲覆盖圆和抛物线,且仅以代数对象的方式(圆方程、顶点式二次函数)出现,以焦点 / 准线轨迹的方式出现。椭圆、双曲线、离心率以及一般二次式的判别式测试(§5)在三省核心课纲中均不存在。下面的八行表告诉你当前大纲下应学哪些节。

Curriculum gap notice (Canadian students).课纲空缺提示(加拿大学生)。 Sections 3, 4, 5, and 7 of this guide are honors / AP-feeder territory for Ontario MCR3U / MHF4U, BC PC 11 / PC 12, and Alberta Math 20-1 / 30-1. Those provincial curricula do not list ellipses, hyperbolas, focus / directrix definitions, the $B^{2} - 4AC$ discriminant test, or planetary-orbit / hyperbolic-navigation applications in the published Grades 10-12 expectations. We've extracted directly from math_grades_11-12_extract.md, pc11_elab_extract.md, pc12_elab_extract.md, and pos_10-12_indicators.txt , the words "ellipse", "hyperbola", "directrix", "eccentricity" do not appear. If you are a Canadian student preparing for AP Calculus, AP Pre-Calc, or IB Math AA HL, work the whole guide. If you are tracking your provincial Grade 12 exit exam only, §1 and §2 plus the circle / parabola subsections of §6 will suffice.本指南 §3、§4、§5、§7 对安大略 MCR3U / MHF4U、BC PC 11 / PC 12、阿尔伯塔 Math 20-1 / 30-1 而言均属荣誉 / AP 衔接内容。这些省级课纲在已发布的 10-12 年级期望中均未列出椭圆、双曲线、焦点 / 准线定义、$B^{2} - 4AC$ 判别式测试,或行星轨道 / 双曲线导航等应用。我们已从 math_grades_11-12_extract.mdpc11_elab_extract.mdpc12_elab_extract.mdpos_10-12_indicators.txt 直接提取核查 , "ellipse"、"hyperbola"、"directrix"、"eccentricity" 等词均未出现。如果你是加拿大学生且备战 AP Calculus、AP Pre-Calc 或 IB Math AA HL,请通读全文;如果你只针对省级 12 年级出口考试,§1、§2 加上 §6 的圆 / 抛物线部分即可。
If you are in…如果你在… Focus on these sections重点学习 Defer / skip可推迟 Source依据
🇺🇸 US Grade 10 (Geometry) , HSG-GPE.A美国 10 年级(几何), HSG-GPE.A §1 (circle , .A.1), §2 (parabola , .A.2), §6 (translations)§1(圆 , .A.1)、§2(抛物线 , .A.2)、§6(平移) §3, §4 (ellipse / hyperbola are HSG-GPE.A.3, marked (+) honors), §5 (discriminant), §7 (applications , not standard-cited)§3、§4(椭圆 / 双曲线属 HSG-GPE.A.3,标 (+) 荣誉级)、§5(判别式)、§7(应用 , 非标准引用) ccssm_hs_math.pdf , HSG-GPE.A.1 (circle, complete-the-square), .A.2 (parabola from focus and directrix), .A.3 (+) (ellipses and hyperbolas from foci), HSG-GPE.A.1(圆、配方法)、.A.2(焦点 / 准线下的抛物线)、.A.3 (+)(焦点下的椭圆与双曲线)
🇨🇦 ON Grade 10 , MPM2D安大略 10 年级 , MPM2D §1 (circle equation centred at origin only) Honors§1(仅以原点为中心的圆方程)Honors §2, §3, §4, §5, §6, §7 , MPM2D treats only the circle at this layer. Parabola work waits for MCR3U as a quadratic function, not a conic.§2、§3、§4、§5、§6、§7 , MPM2D 此层仅处理圆。抛物线留到 MCR3U,且作为二次函数而非圆锥曲线。 math_grades_9-10.pdf , MPM2D Analytic Geometry strand covers $x^{2} + y^{2} = r^{2}$ at the origin. Translated circles and named conics are not listed in the 9-10 document., MPM2D 解析几何单元覆盖原点处的 $x^{2} + y^{2} = r^{2}$。9-10 文件未列平移圆与命名圆锥曲线。
🇨🇦 ON Grade 11-12 , MCR3U / MHF4U安大略 11-12 年级 , MCR3U / MHF4U §1, §2 (parabola as vertex-form quadratic), §6 (translations of $y = a(x-h)^{2} + k$) Honors§1、§2(顶点式抛物线)、§6($y = a(x-h)^{2} + k$ 的平移)Honors §3, §4, §5, §7 , MCR3U Strand A treats parabolas as quadratic functions only; ellipse / hyperbola / directrix / focus language does not appear. Treat as enrichment for AP-feeder routes.§3、§4、§5、§7 , MCR3U 单元 A 仅把抛物线作为二次函数;不出现椭圆 / 双曲线 / 准线 / 焦点等术语。AP 衔接方向上视为拓展。 math_grades_11-12.pdf , MCR3U Strand A Quadratic Functions; no occurrence of "ellipse" or "hyperbola" in the extract., MCR3U 单元 A 二次函数;提取文档中未出现"ellipse"或"hyperbola"。
🇨🇦 BC Grade 11 , PC 11BC 11 年级 , PC 11 §1 (circle equation), §2 (parabola as quadratic vertex form), §6 (vertex shifts) Honors§1(圆方程)、§2(顶点式抛物线)、§6(顶点平移)Honors §3, §4, §5, §7 , PC 11 elaborations name only quadratic functions / equations as parabola-bearing content; no ellipse / hyperbola entries.§3、§4、§5、§7 , PC 11 详释仅列二次函数 / 方程作为抛物线相关内容;无椭圆 / 双曲线条目。 pc11_elab.pdf , Content: "quadratic functions and equations" (vertex form). The extract contains no occurrence of ellipse / hyperbola / directrix., 内容:"二次函数与方程"(顶点式)。提取文档中无 ellipse / hyperbola / directrix。
🇨🇦 BC Grade 12 , PC 12BC 12 年级 , PC 12 §1, §2, §6 (translations of functions / relations). Treat §3-§5 as AP-feeder enrichment for university entry. Honors§1、§2、§6(函数 / 关系的变换)。§3-§5 视为面向大学入学的 AP 衔接拓展。Honors §3, §4, §5, §7 , the extracted PC 12 Big Ideas and content list do not name ellipses / hyperbolas / directrix / eccentricity.§3、§4、§5、§7 , PC 12 提取的大概念与内容列表未提及椭圆 / 双曲线 / 准线 / 离心率。 pc12_elab.pdf , Content: "transformations of functions and relations." No ellipse / hyperbola in the extract., 内容:"函数与关系的变换"。提取文档中无椭圆 / 双曲线。
🇨🇦 AB Grade 10 , Math 10C阿尔伯塔 10 年级 , Math 10C §1 (circle as 2-D figure; algebraic equation foreshadowed) only.仅 §1(圆作为二维图形;代数方程作铺垫)。 §2 through §7 , Math 10C does not formalise the conic equations; parabola work waits for Math 20-1.§2 至 §7 , Math 10C 未正式化圆锥曲线方程;抛物线留待 Math 20-1。 pos_10-12_indicators.pdf , Math 10C Measurement and RF general outcomes; no specific outcome naming parabola / ellipse / hyperbola., Math 10C 测量与 RF 总目标;无具体目标提及抛物线 / 椭圆 / 双曲线。
🇨🇦 AB Grade 11-12 , Math 20-1 / 30-1阿尔伯塔 11-12 年级 , Math 20-1 / 30-1 §1, §2 (parabola as $y = a(x-p)^{2} + q$, Math 20-1 RF GO 3, indicators 3.1-3.9), §6 (vertex translations) Honors§1、§2(抛物线作为 $y = a(x-p)^{2} + q$,Math 20-1 RF 总目标 3,指标 3.1-3.9)、§6(顶点平移)Honors §3, §4, §5, §7 , Math 20-1 RF GO 3 treats the parabola as a quadratic function only. The POS extract has no specific outcome on ellipse / hyperbola / focus / directrix / eccentricity at any grade.§3、§4、§5、§7 , Math 20-1 RF 总目标 3 仅以二次函数处理抛物线。POS 提取在所有年级均无关于椭圆 / 双曲线 / 焦点 / 准线 / 离心率的具体目标。 pos_10-12_indicators.pdf , Math 20-1 RF General Outcome 3, specific outcomes 3.1-3.9 (analyse $y = a(x-p)^{2} + q$). Conic-section terminology is not present in the POS extract., Math 20-1 RF 总目标 3,具体目标 3.1-3.9(分析 $y = a(x-p)^{2} + q$)。POS 提取中没有圆锥曲线术语。
🇺🇸 US AP-feeder (Pre-Calc / Honors)美国 AP 衔接(Pre-Calc / 荣誉) All seven sections, with depth on §3-§5. AP Pre-Calc and IB Math AA HL both assume fluent conic identification and the focus / directrix locus definitions.全部 7 节,§3-§5 加深。AP Pre-Calc 与 IB Math AA HL 均默认你能熟练辨识圆锥曲线并掌握焦点 / 准线轨迹定义。 Nothing , treat this as the geometric scaffold for parametric / polar curves and for orbital-mechanics applications later.无 , 视为参数 / 极坐标曲线及后续轨道力学应用的几何骨架。 ccssm_hs_math.pdf , (+) cluster HSG-GPE.A.3 is explicitly the AP-feeder content; AP Pre-Calc CED includes conic equations under Unit 3 (Trigonometric and Polar Functions) and Unit 4 (Functions Involving Parameters and Vectors)., (+) 簇 HSG-GPE.A.3 正是 AP 衔接内容;AP Pre-Calc 课纲在 Unit 3(三角与极坐标函数)与 Unit 4(含参数与向量的函数)中包含圆锥曲线方程。

Once you have located your row, use the two cards below for the speed at which you should work through the recommended sections.找到所在行后,用下面两张卡片决定推进速度。

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If you are cramming the night before如果你在临阵磨枪

Memorise four equations and one test. Circle: $(x-h)^{2} + (y-k)^{2} = r^{2}$. Vertical-axis parabola: $y - k = \frac{1}{4p}(x-h)^{2}$ with focus $(h, k+p)$ and directrix $y = k - p$. Ellipse: $\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1$ with $c^{2} = a^{2} - b^{2}$. Hyperbola: $\frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1$ with $c^{2} = a^{2} + b^{2}$ and asymptotes $y - k = \pm \frac{b}{a}(x - h)$. The test: discriminant $B^{2} - 4AC$ on the general form tells you which conic ($<0$ ellipse, $=0$ parabola, $>0$ hyperbola).背熟四个方程与一个判别式。圆:$(x-h)^{2} + (y-k)^{2} = r^{2}$。竖轴抛物线:$y - k = \frac{1}{4p}(x-h)^{2}$,焦点 $(h, k+p)$,准线 $y = k - p$。椭圆:$\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1$,$c^{2} = a^{2} - b^{2}$。双曲线:$\frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1$,$c^{2} = a^{2} + b^{2}$,渐近线 $y - k = \pm \frac{b}{a}(x - h)$。判别式:一般式中的 $B^{2} - 4AC$ 给出曲线类型($<0$ 椭圆,$=0$ 抛物线,$>0$ 双曲线)。

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If you are going for the top mark如果你目标顶分

Always derive the standard form from the locus definition once before relying on it: the parabola from $\mathrm{dist}(P, \text{focus}) = \mathrm{dist}(P, \text{directrix})$; the ellipse from $|PF_{1}| + |PF_{2}| = 2a$; the hyperbola from $||PF_{1}| - |PF_{2}|| = 2a$. Always complete the square twice on the general form before reading off the centre. For an ellipse / hyperbola, work out eccentricity $e = c / a$ and learn what $e \to 0$, $e = 1$, $e > 1$ each mean , the unifying parameter behind all four conics.用前先从轨迹定义至少推一次标准式:抛物线由 $\mathrm{dist}(P, \text{焦点}) = \mathrm{dist}(P, \text{准线})$;椭圆由 $|PF_{1}| + |PF_{2}| = 2a$;双曲线由 $||PF_{1}| - |PF_{2}|| = 2a$。一般式中务必两次配方后再读中心。椭圆 / 双曲线再算离心率 $e = c / a$,并理解 $e \to 0$、$e = 1$、$e > 1$ 各自含义 , 这是把四种曲线统一起来的参数。

Honors flag.荣誉级标记。 Sections §3 (Ellipses), §4 (Hyperbolas), §5 (Conic discriminant), and §7 (Applications) carry the Honors chip for Ontario, BC, and Alberta. None of the provincial Grades 10-12 documents extract a locus-based ellipse or hyperbola definition; none names the $B^{2} - 4AC$ test by name. US Geometry students see §3-§5 as the (+) tier of HSG-GPE.A.3, and AP-feeder Pre-Calc treats all of §1-§7 as core. If your row sends you to these sections, treat them as required content for AP / IB tracks, not as enrichment.§3(椭圆)、§4(双曲线)、§5(圆锥曲线判别式)与 §7(应用)在安大略、BC、阿尔伯塔均标 Honors。三省 10-12 年级文件均未提取出基于轨迹的椭圆 / 双曲线定义;均未点名 $B^{2} - 4AC$ 判别式。美国几何课将 §3-§5 视为 HSG-GPE.A.3 的 (+) 层,AP 衔接 Pre-Calc 将 §1-§7 全部视为核心。如果你的行指向上述节,且走 AP / IB 方向,请按必学处理,不是拓展。

Distance, Midpoint, and the Equation of a Circle距离、中点与圆方程

Three formulae the rest of conics is built on.圆锥曲线其余内容的三个基石公式。 $$ d \;=\; \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}, \qquad M \;=\; \left(\tfrac{x_{1} + x_{2}}{2}, \; \tfrac{y_{1} + y_{2}}{2}\right), \qquad (x - h)^{2} + (y - k)^{2} \;=\; r^{2}. $$
  • Distance.距离公式。 Direct application of the Pythagorean theorem to the right triangle whose legs are the differences in coordinates.把坐标差作直角边,对相应直角三角形应用勾股定理即得。
  • Midpoint.中点公式。 Coordinate averages. Useful for finding centres of circles when two endpoints of a diameter are known.坐标分量平均。已知直径两端点时用以求圆心。
  • Circle.圆方程。 $(x-h)^{2} + (y-k)^{2} = r^{2}$ , the locus of points at distance $r$ from centre $(h, k)$. This is CCSSM HSG-GPE.A.1 verbatim , the standard explicitly names "deriving the circle equation via the Pythagorean theorem" as the expected approach.$(x-h)^{2} + (y-k)^{2} = r^{2}$ , 到圆心 $(h, k)$ 距离为 $r$ 的点的轨迹。这正是 CCSSM HSG-GPE.A.1,标准明确以"由勾股定理推导圆方程"作为期望路径。
  • General form.一般形式。 $x^{2} + y^{2} + Dx + Ey + F = 0$ is a circle iff the implied radius $r = \tfrac{1}{2}\sqrt{D^{2} + E^{2} - 4F}$ is real and positive. Recover centre / radius by completing the square in $x$ and $y$ separately.$x^{2} + y^{2} + Dx + Ey + F = 0$ 为圆当且仅当隐含半径 $r = \tfrac{1}{2}\sqrt{D^{2} + E^{2} - 4F}$ 为正实数。分别对 $x$、$y$ 配方即可读出圆心 / 半径。
Worked Example 1 · From general form to centre + radius例题 1 · 由一般式求圆心与半径

Find the centre and radius of the circle $x^{2} + y^{2} - 6x + 10y + 9 = 0$.求圆 $x^{2} + y^{2} - 6x + 10y + 9 = 0$ 的圆心与半径。

Group $x$ and $y$ separately, then complete the square.分别整合 $x$ 与 $y$ 项,再配方。

$$ (x^{2} - 6x) + (y^{2} + 10y) = -9. $$ $$ (x^{2} - 6x + 9) + (y^{2} + 10y + 25) = -9 + 9 + 25 = 25. $$

Read off the standard form.读出标准形式。

$$ (x - 3)^{2} + (y + 5)^{2} = 25 \quad\Longrightarrow\quad \text{centre } (3, -5), \;\; \text{radius } 5. $$

Sanity-check.合理性核验。 The point on the circle directly above the centre is $(3, 0)$; substitute into the original equation: $9 + 0 - 18 + 0 + 9 = 0$ ✓.圆心正上方的圆上点 $(3, 0)$ 代入原方程:$9 + 0 - 18 + 0 + 9 = 0$ ✓。

Find the equation of the circle with centre $(-2, 4)$ passing through $(1, 8)$.求圆心 $(-2, 4)$ 且过 $(1, 8)$ 的圆方程。
§1 · Q1
$(x+2)^{2} + (y-4)^{2} = 5$
$(x-2)^{2} + (y+4)^{2} = 25$
$(x+2)^{2} + (y-4)^{2} = 25$
$(x-1)^{2} + (y-8)^{2} = 25$
$r^{2} = (1-(-2))^{2} + (8 - 4)^{2} = 9 + 16 = 25$. Standard form: $(x - (-2))^{2} + (y - 4)^{2} = 25$, i.e. $(x+2)^{2} + (y-4)^{2} = 25$.$r^{2} = (1-(-2))^{2} + (8 - 4)^{2} = 9 + 16 = 25$。标准式:$(x - (-2))^{2} + (y - 4)^{2} = 25$,即 $(x+2)^{2} + (y-4)^{2} = 25$。
Distance-squared from centre to the given point is $r^{2}$. Then write $(x - h)^{2} + (y - k)^{2} = r^{2}$ with $(h, k) = (-2, 4)$.圆心到给定点的距离平方即 $r^{2}$。再以 $(h, k) = (-2, 4)$ 写 $(x - h)^{2} + (y - k)^{2} = r^{2}$。
Find the centre of the circle $x^{2} + y^{2} + 4x - 8y - 5 = 0$.求圆 $x^{2} + y^{2} + 4x - 8y - 5 = 0$ 的圆心。
§1 · Q2
$(4, -8)$
$(-2, 4)$
$(2, -4)$
$(-4, 8)$
Complete the square: $(x^{2} + 4x + 4) + (y^{2} - 8y + 16) = 5 + 4 + 16 = 25$, giving $(x + 2)^{2} + (y - 4)^{2} = 25$. Centre $(-2, 4)$, radius $5$.配方:$(x^{2} + 4x + 4) + (y^{2} - 8y + 16) = 5 + 4 + 16 = 25$,得 $(x + 2)^{2} + (y - 4)^{2} = 25$。圆心 $(-2, 4)$,半径 $5$。
Centre coordinates from $x^{2} + y^{2} + Dx + Ey + F = 0$ are $\left(-\tfrac{D}{2}, -\tfrac{E}{2}\right)$. Here $D = 4$ and $E = -8$, so centre $(-2, 4)$.由 $x^{2} + y^{2} + Dx + Ey + F = 0$ 直接得圆心 $\left(-\tfrac{D}{2}, -\tfrac{E}{2}\right)$。本题 $D = 4$、$E = -8$,故圆心 $(-2, 4)$。

Parabolas: Vertex, Focus, Directrix抛物线:顶点、焦点、准线

The locus definition and the two standard forms.轨迹定义与两种标准式。

A parabola is the locus of points equidistant from a fixed point (the focus, $F$) and a fixed line (the directrix). CCSSM HSG-GPE.A.2 names this definition explicitly and asks students to derive the equation from it.抛物线是与一固定点(焦点 $F$)和一固定直线(准线 directrix)等距的点的轨迹。CCSSM HSG-GPE.A.2 正是以此定义为基础,要求学生由它推导方程。

$$ \text{Vertical axis: } \;\; y - k \;=\; \frac{1}{4p}(x - h)^{2}, \qquad \text{Horizontal axis: } \;\; x - h \;=\; \frac{1}{4p}(y - k)^{2}. $$
  • Vertical-axis parts.竖轴抛物线各要素。 Vertex $(h, k)$, focus $(h, k + p)$, directrix $y = k - p$. Opens up if $p > 0$, down if $p < 0$.顶点 $(h, k)$,焦点 $(h, k + p)$,准线 $y = k - p$。$p > 0$ 开口向上,$p < 0$ 开口向下。
  • Horizontal-axis parts.横轴抛物线各要素。 Vertex $(h, k)$, focus $(h + p, k)$, directrix $x = h - p$. Opens right if $p > 0$, left if $p < 0$.顶点 $(h, k)$,焦点 $(h + p, k)$,准线 $x = h - p$。$p > 0$ 开口向右,$p < 0$ 开口向左。
  • Vertex form.顶点式。 $y = a(x - h)^{2} + k$ with $a = \frac{1}{4p}$ is the same equation written in function-of-$x$ form , this is the version Ontario MCR3U / BC PC 11 / AB Math 20-1 use.$y = a(x - h)^{2} + k$(其中 $a = \frac{1}{4p}$)是同一方程以 $x$ 的函数形式书写 , 安大略 MCR3U / BC PC 11 / 阿尔伯塔 Math 20-1 都采用此版本。
Worked Example 2 · Parabola from focus and directrix例题 2 · 由焦点与准线求抛物线

Find the standard-form equation of the parabola with focus $(2, 5)$ and directrix $y = 1$.求焦点 $(2, 5)$、准线 $y = 1$ 的抛物线的标准方程。

Find the vertex.求顶点。 The vertex sits midway between the focus and the directrix, on the axis perpendicular to the directrix. Vertex $= (2, \tfrac{5 + 1}{2}) = (2, 3)$.顶点位于焦点与准线之间,且在垂直于准线的轴上。顶点 $= (2, \tfrac{5 + 1}{2}) = (2, 3)$。

Find $p$.求 $p$。 Signed distance from vertex to focus along the axis: $p = 5 - 3 = 2$. (Positive, so the parabola opens up.)沿轴线由顶点到焦点的有向距离:$p = 5 - 3 = 2$。(为正,开口向上。)

Write the equation.写出方程。

$$ y - 3 \;=\; \frac{1}{4 \cdot 2}(x - 2)^{2} \;=\; \frac{1}{8}(x - 2)^{2}. $$

Sanity-check at the vertex and one other point.在顶点与另一点处核验。 Vertex $(2, 3)$: distance to focus $= 2$, distance to directrix $y = 1$ also $= 2$. ✓ Point $(6, 5)$ on the parabola? $5 - 3 = \tfrac{1}{8}(4)^{2} = 2$. ✓ And $(6, 5)$ has distance $\sqrt{16 + 0} = 4$ to the focus and distance $5 - 1 = 4$ to the directrix , the locus property holds.顶点 $(2, 3)$:到焦点距离 $= 2$,到准线 $y = 1$ 距离也 $= 2$ ✓。点 $(6, 5)$ 在抛物线上?$5 - 3 = \tfrac{1}{8}(4)^{2} = 2$ ✓。且 $(6, 5)$ 到焦点距离 $\sqrt{16 + 0} = 4$,到准线距离 $5 - 1 = 4$ , 轨迹性质成立。

A parabola has vertex $(0, 0)$ and focus $(0, 3)$. Find its standard equation.抛物线顶点 $(0, 0)$、焦点 $(0, 3)$。求其标准方程。
§2 · Q1
$y = \tfrac{1}{12} x^{2}$
$y = 12 x^{2}$
$x = \tfrac{1}{12} y^{2}$
$y = \tfrac{1}{3} x^{2}$
Focus above vertex $\Rightarrow$ vertical axis, opens up. $p = 3 \Rightarrow$ coefficient $= \tfrac{1}{4p} = \tfrac{1}{12}$. So $y = \tfrac{1}{12} x^{2}$.焦点在顶点正上方 $\Rightarrow$ 竖轴、开口向上。$p = 3 \Rightarrow$ 系数 $= \tfrac{1}{4p} = \tfrac{1}{12}$,故 $y = \tfrac{1}{12} x^{2}$。
For a parabola opening up, $y - k = \tfrac{1}{4p}(x - h)^{2}$. Here vertex is origin and $p = 3$, so $y = x^{2}/(4p) = x^{2}/12$.开口向上时 $y - k = \tfrac{1}{4p}(x - h)^{2}$。顶点为原点,$p = 3$,故 $y = x^{2}/(4p) = x^{2}/12$。
For the parabola $y = -\tfrac{1}{8}(x - 4)^{2} + 2$, find the focus.对抛物线 $y = -\tfrac{1}{8}(x - 4)^{2} + 2$,求焦点。
§2 · Q2
$(4, 4)$
$(4, 2)$
$(4, 0)$
$(2, 4)$
Vertex $(4, 2)$. Coefficient $\tfrac{1}{4p} = -\tfrac{1}{8}$ gives $p = -2$ (negative $\Rightarrow$ opens down). Focus $= (h, k + p) = (4, 2 + (-2)) = (4, 0)$.顶点 $(4, 2)$。系数 $\tfrac{1}{4p} = -\tfrac{1}{8}$ 得 $p = -2$(为负 $\Rightarrow$ 开口向下)。焦点 $= (h, k + p) = (4, 2 + (-2)) = (4, 0)$。
Read $4p$ from the reciprocal of the leading coefficient; sign of $p$ tells the direction of opening. Focus sits at $(h, k + p)$.由首项系数的倒数求 $4p$;$p$ 的正负决定开口方向。焦点为 $(h, k + p)$。

Ellipses: Two Foci, Sum of Distances Honors — ON / BC / AB椭圆:两焦点、距离之和 荣誉 — ON / BC / AB

Curriculum note.课纲提示。 Ellipses are not in the published Grades 10-12 curricula of Ontario, BC, or Alberta. This section is core for US Common Core HSG-GPE.A.3 ((+) honors), AP Pre-Calc, and IB Math AA HL B1.椭圆不在安、BC、阿三省 10-12 年级已发布的课纲中。本节对美国 CCSSM HSG-GPE.A.3((+) 荣誉级)、AP Pre-Calc 与 IB Math AA HL B1 而言均为核心。
Locus definition and the standard equation.轨迹定义与标准方程。

An ellipse is the locus of points $P$ such that $|PF_{1}| + |PF_{2}| = 2a$, where $F_{1}$ and $F_{2}$ are the two foci. The standard form (centred at origin, major axis horizontal) is:椭圆是满足 $|PF_{1}| + |PF_{2}| = 2a$ 的点 $P$ 的轨迹,$F_{1}$、$F_{2}$ 为两焦点。以原点为中心、主轴水平的标准方程为:

$$ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} \;=\; 1 \qquad \text{with } a > b > 0, \quad c^{2} \;=\; a^{2} - b^{2}. $$
  • Vertices.主顶点。 $(\pm a, 0)$ on the major axis.主轴方向上的 $(\pm a, 0)$。
  • Co-vertices.短轴端点。 $(0, \pm b)$ on the minor axis.短轴方向上的 $(0, \pm b)$。
  • Foci.焦点。 $(\pm c, 0)$ on the major axis, where $c = \sqrt{a^{2} - b^{2}}$.主轴方向上的 $(\pm c, 0)$,其中 $c = \sqrt{a^{2} - b^{2}}$。
  • Eccentricity.离心率。 $$e = \tfrac{c}{a}, \quad 0 \le e < 1.$$ $e = 0$ is a circle; $e \to 1$ is a very elongated ellipse.$e = 0$ 即圆;$e \to 1$ 椭圆极扁长。
  • Vertical major axis.主轴竖直情形。 If $b > a$, swap roles: $\tfrac{x^{2}}{b^{2}} + \tfrac{y^{2}}{a^{2}} = 1$ with $a^{2}$ now under $y^{2}$ and foci on the $y$-axis. (Some texts always reserve $a$ for the longer semi-axis , we follow that convention.)若 $b > a$,互换:$\tfrac{x^{2}}{b^{2}} + \tfrac{y^{2}}{a^{2}} = 1$,$a^{2}$ 现在在 $y^{2}$ 下方,焦点在 $y$ 轴。(部分教材始终用 $a$ 表示更长半轴 , 本指南采用该约定。)
Worked Example 3 · Parts of an ellipse from its equation例题 3 · 由方程求椭圆各要素

For the ellipse $\tfrac{x^{2}}{25} + \tfrac{y^{2}}{9} = 1$, find the vertices, co-vertices, foci, and eccentricity.对椭圆 $\tfrac{x^{2}}{25} + \tfrac{y^{2}}{9} = 1$,求主顶点、短轴端点、焦点与离心率。

Identify $a$ and $b$.识别 $a$、$b$。 $a^{2} = 25 \Rightarrow a = 5$ (the larger one, denominator under $x^{2}$ , so the major axis is horizontal). $b^{2} = 9 \Rightarrow b = 3$.$a^{2} = 25 \Rightarrow a = 5$(较大者,在 $x^{2}$ 下 , 故主轴水平)。$b^{2} = 9 \Rightarrow b = 3$。

Vertices and co-vertices.主顶点与短轴端点。

$$ \text{Vertices: } (\pm 5, 0). \qquad \text{Co-vertices: } (0, \pm 3). $$

Compute $c$ and the foci.求 $c$ 与焦点。

$$ c \;=\; \sqrt{a^{2} - b^{2}} \;=\; \sqrt{25 - 9} \;=\; 4. \qquad \text{Foci: } (\pm 4, 0). $$

Eccentricity.离心率。 $e = c/a = 4/5 = 0.8$. Moderately elongated.椭圆中等扁长。

Locus check.轨迹核验。 Take the vertex $(5, 0)$: $|PF_{1}| + |PF_{2}| = (5 - (-4)) + (5 - 4) = 9 + 1 = 10 = 2a$. ✓取主顶点 $(5, 0)$:$|PF_{1}| + |PF_{2}| = (5 - (-4)) + (5 - 4) = 9 + 1 = 10 = 2a$ ✓。

An ellipse centred at the origin has vertices $(\pm 6, 0)$ and passes through $(0, 2)$. Find its standard-form equation.椭圆中心在原点,主顶点 $(\pm 6, 0)$ 并过 $(0, 2)$。求标准方程。
§3 · Q1
$\tfrac{x^{2}}{6} + \tfrac{y^{2}}{2} = 1$
$\tfrac{x^{2}}{36} + \tfrac{y^{2}}{4} = 1$
$\tfrac{x^{2}}{4} + \tfrac{y^{2}}{36} = 1$
$\tfrac{x^{2}}{36} - \tfrac{y^{2}}{4} = 1$
$a = 6 \Rightarrow a^{2} = 36$. The point $(0, 2)$ on the minor axis gives $b = 2$, so $b^{2} = 4$. Major axis horizontal, so $\tfrac{x^{2}}{36} + \tfrac{y^{2}}{4} = 1$.$a = 6 \Rightarrow a^{2} = 36$。点 $(0, 2)$ 在短轴上给 $b = 2$,故 $b^{2} = 4$。主轴水平,故 $\tfrac{x^{2}}{36} + \tfrac{y^{2}}{4} = 1$。
Vertices give $a$; the perpendicular intercept gives $b$. Plug into $\tfrac{x^{2}}{a^{2}} + \tfrac{y^{2}}{b^{2}} = 1$ with major axis along the longer semi-axis.主顶点给 $a$;垂直方向截距给 $b$。代入 $\tfrac{x^{2}}{a^{2}} + \tfrac{y^{2}}{b^{2}} = 1$,主轴沿较长半轴方向。
For the ellipse $\tfrac{x^{2}}{16} + \tfrac{y^{2}}{25} = 1$, find the eccentricity.对椭圆 $\tfrac{x^{2}}{16} + \tfrac{y^{2}}{25} = 1$,求离心率。
§3 · Q2
$3/4$
$4/5$
$3/5$
$5/3$
Major axis vertical (larger denominator under $y^{2}$): $a^{2} = 25 \Rightarrow a = 5$; $b^{2} = 16 \Rightarrow b = 4$. $c = \sqrt{a^{2} - b^{2}} = \sqrt{9} = 3$. $e = c/a = 3/5$.主轴竖直(较大分母在 $y^{2}$ 下):$a^{2} = 25 \Rightarrow a = 5$,$b^{2} = 16 \Rightarrow b = 4$。$c = \sqrt{a^{2} - b^{2}} = 3$。$e = c/a = 3/5$。
For an ellipse, the larger of the two squared denominators is $a^{2}$. Then $c = \sqrt{a^{2} - b^{2}}$ and $e = c/a$.对椭圆,两分母平方中较大者为 $a^{2}$。再求 $c = \sqrt{a^{2} - b^{2}}$ 与 $e = c/a$。

Hyperbolas: Two Foci, Difference of Distances Honors — ON / BC / AB双曲线:两焦点、距离之差 荣誉 — ON / BC / AB

Curriculum note.课纲提示。 Hyperbolas are not in the published Grades 10-12 curricula of Ontario, BC, or Alberta. This section is core for US Common Core HSG-GPE.A.3 ((+) honors), AP Pre-Calc, and IB Math AA HL.双曲线不在安、BC、阿三省 10-12 年级已发布的课纲中。本节对美国 CCSSM HSG-GPE.A.3((+) 荣誉级)、AP Pre-Calc 与 IB Math AA HL 而言均为核心。
Locus definition and the standard equation.轨迹定义与标准方程。

A hyperbola is the locus of points $P$ such that $\big| |PF_{1}| - |PF_{2}| \big| = 2a$ , the absolute difference of distances to two foci is constant. The standard form (centred at origin, transverse axis horizontal) is:双曲线是满足 $\big| |PF_{1}| - |PF_{2}| \big| = 2a$ 的点 $P$ 的轨迹 , 到两焦点距离之(绝对值)为常数。以原点为中心、实轴水平的标准方程为:

$$ \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} \;=\; 1, \qquad c^{2} \;=\; a^{2} + b^{2}. $$
  • Vertices.顶点。 $(\pm a, 0)$ on the transverse axis. The two branches open outward through these vertices.实轴上的 $(\pm a, 0)$。两支由这两顶点向外延展。
  • Foci.焦点。 $(\pm c, 0)$ with $c = \sqrt{a^{2} + b^{2}}$ , note the plus sign, opposite to the ellipse.$(\pm c, 0)$,$c = \sqrt{a^{2} + b^{2}}$ , 注意是加号,与椭圆相反。
  • Asymptotes.渐近线。 $$y = \pm \tfrac{b}{a} x.$$ The branches approach but never touch these lines as $|x| \to \infty$.$|x| \to \infty$ 时两支无限趋近但不触及。
  • Eccentricity.离心率。 $e = c/a > 1.$ Always greater than $1$ , that's what separates hyperbolas from ellipses on the eccentricity scale.总大于 $1$ , 这正是离心率刻度上双曲线与椭圆的分界。
  • Vertical transverse axis.实轴竖直情形。 $$\tfrac{y^{2}}{a^{2}} - \tfrac{x^{2}}{b^{2}} = 1$$ with vertices on the $y$-axis at $(0, \pm a)$ and asymptotes $y = \pm \tfrac{a}{b} x$.顶点 $(0, \pm a)$,渐近线 $y = \pm \tfrac{a}{b} x$。
The asymptote rectangle. Sketch the rectangle with corners $(\pm a, \pm b)$. The asymptotes are its diagonals; the vertices sit at the midpoints of two opposite sides. This box is the fastest way to draw a hyperbola.渐近矩形。 画四角为 $(\pm a, \pm b)$ 的矩形。渐近线是其对角线;顶点位于两条对边中点。这是画双曲线最快的方法。
Worked Example 4 · Parts of a hyperbola from its equation例题 4 · 由方程求双曲线各要素

For the hyperbola $\tfrac{x^{2}}{9} - \tfrac{y^{2}}{16} = 1$, find the vertices, foci, asymptotes, and eccentricity.对双曲线 $\tfrac{x^{2}}{9} - \tfrac{y^{2}}{16} = 1$,求顶点、焦点、渐近线与离心率。

Identify $a$ and $b$.识别 $a$、$b$。 $a^{2} = 9 \Rightarrow a = 3$; $b^{2} = 16 \Rightarrow b = 4$. Note that for a hyperbola $a$ is the denominator of the positive term, not necessarily the larger one.$a^{2} = 9 \Rightarrow a = 3$;$b^{2} = 16 \Rightarrow b = 4$。对双曲线,$a$ 是正项的分母,不必是较大者。

Vertices and asymptotes.顶点与渐近线。

$$ \text{Vertices: } (\pm 3, 0). \qquad \text{Asymptotes: } y = \pm \tfrac{4}{3} x. $$

Compute $c$ and the foci.求 $c$ 与焦点。

$$ c \;=\; \sqrt{a^{2} + b^{2}} \;=\; \sqrt{9 + 16} \;=\; 5. \qquad \text{Foci: } (\pm 5, 0). $$

Eccentricity.离心率。 $e = c/a = 5/3 \approx 1.67.$

Locus check.轨迹核验。 Take the vertex $(3, 0)$: $\big| |PF_{1}| - |PF_{2}| \big| = |(3 - (-5)) - (3 - 5)| = |8 - (-2)| / ... $ , more cleanly, distances are $|3 + 5| = 8$ and $|3 - 5| = 2$; difference $= 6 = 2a$. ✓取顶点 $(3, 0)$:到两焦点距离分别为 $|3 + 5| = 8$ 与 $|3 - 5| = 2$;差 $= 6 = 2a$ ✓。

Identify the foci of the hyperbola $\tfrac{y^{2}}{16} - \tfrac{x^{2}}{9} = 1$.求双曲线 $\tfrac{y^{2}}{16} - \tfrac{x^{2}}{9} = 1$ 的焦点。
§4 · Q1
$(\pm 5, 0)$
$(\pm 4, 0)$
$(0, \pm 5)$
$(0, \pm \sqrt{7})$
Positive term is $y^{2}$ $\Rightarrow$ transverse axis is vertical. $a^{2} = 16$, $b^{2} = 9$. $c^{2} = a^{2} + b^{2} = 25 \Rightarrow c = 5$. Foci $(0, \pm 5)$.正项为 $y^{2}$ $\Rightarrow$ 实轴竖直。$a^{2} = 16$,$b^{2} = 9$。$c^{2} = a^{2} + b^{2} = 25 \Rightarrow c = 5$。焦点 $(0, \pm 5)$。
The positive term tells the transverse axis direction. For a hyperbola, $c^{2} = a^{2} + b^{2}$ (sum, not difference).正项的位置决定实轴方向。对双曲线 $c^{2} = a^{2} + b^{2}$(和,非差)。
Find the asymptotes of $\tfrac{x^{2}}{4} - \tfrac{y^{2}}{9} = 1$.求 $\tfrac{x^{2}}{4} - \tfrac{y^{2}}{9} = 1$ 的渐近线。
§4 · Q2
$y = \pm \tfrac{3}{2} x$
$y = \pm \tfrac{2}{3} x$
$y = \pm 6 x$
$y = \pm \tfrac{1}{2} x$
$a^{2} = 4 \Rightarrow a = 2$, $b^{2} = 9 \Rightarrow b = 3$. Horizontal transverse axis $\Rightarrow$ asymptotes $y = \pm \tfrac{b}{a} x = \pm \tfrac{3}{2} x$.$a^{2} = 4 \Rightarrow a = 2$,$b^{2} = 9 \Rightarrow b = 3$。实轴水平 $\Rightarrow$ 渐近线 $y = \pm \tfrac{b}{a} x = \pm \tfrac{3}{2} x$。
For horizontal transverse axis, asymptotes are $y = \pm (b/a) x$ , ratio of the $y$-denominator's square root over the $x$-denominator's square root.实轴水平时渐近线为 $y = \pm (b/a) x$ , $y$ 分母平方根除以 $x$ 分母平方根。

Identifying a Conic from Its General Form Honors — ON / BC / AB由一般式辨识圆锥曲线 荣誉 — ON / BC / AB

Curriculum note.课纲提示。 The $B^{2} - 4AC$ discriminant test is canonical AP-feeder Pre-Calc / Honors content. CCSSM does not name it at standard level; Ontario, BC, and Alberta do not name it at all in the extracted Grades 10-12 curricula. Treat this section as required only if you're heading to AP Pre-Calc, AP Calculus, or IB Math AA HL.$B^{2} - 4AC$ 判别式测试是 AP 衔接 Pre-Calc / 荣誉级的传统内容。CCSSM 在标准层未点名;安、BC、阿三省 10-12 年级提取课纲完全未提及。本节仅在你走 AP Pre-Calc、AP Calculus 或 IB Math AA HL 路径时为必学。
The general second-degree equation and the discriminant.一般二次方程与判别式。 $$ A x^{2} + B x y + C y^{2} + D x + E y + F \;=\; 0. $$

When this represents a non-degenerate conic (i.e. an actual curve, not a single point / pair of lines / empty set), compute the discriminant $\Delta = B^{2} - 4AC$:当此方程表示非退化的圆锥曲线(即真实曲线,而非单点 / 两直线 / 空集)时,计算判别式 $\Delta = B^{2} - 4AC$:

Sign of $B^{2} - 4AC$$B^{2} - 4AC$ 的符号Conic type曲线类型Special case特殊情形
$< 0$Ellipse椭圆$A = C$, $B = 0$ $\Rightarrow$ circle$A = C$、$B = 0$ $\Rightarrow$ 圆
$= 0$Parabola抛物线degenerate cases: two parallel lines, one line, or empty退化情形:两条平行线、一条线或空集
$> 0$Hyperbola双曲线$B^{2} - 4AC > 0$ and $A + C = 0$ $\Rightarrow$ rectangular hyperbola$B^{2} - 4AC > 0$ $A + C = 0$ $\Rightarrow$ 直角双曲线
  • Without the $xy$ term ($B = 0$).无 $xy$ 项($B = 0$)。 $\Delta = -4AC$. So: $AC > 0$ ellipse / circle; $AC = 0$ (one of $A, C$ is zero) parabola; $AC < 0$ hyperbola.$\Delta = -4AC$。所以 $AC > 0$ 椭圆 / 圆;$AC = 0$($A$、$C$ 之一为 $0$)抛物线;$AC < 0$ 双曲线。
  • When $B \ne 0$.$B \ne 0$ 时。 The conic is rotated. The discriminant still classifies the type; rotating the axes by an angle $\theta$ with $\cot(2\theta) = (A - C)/B$ eliminates the $xy$ term.曲线旋转了。判别式仍能定型;以满足 $\cot(2\theta) = (A - C)/B$ 的角 $\theta$ 旋转坐标轴可消去 $xy$ 项。
Worked Example 5 · Identify the conic例题 5 · 辨识曲线

Classify the conic $4 x^{2} - 9 y^{2} + 16 x + 18 y - 29 = 0$.辨识曲线 $4 x^{2} - 9 y^{2} + 16 x + 18 y - 29 = 0$。

Read off the coefficients.读出系数。 $A = 4$, $B = 0$, $C = -9$.$A = 4$、$B = 0$、$C = -9$。

Compute the discriminant.计算判别式。

$$ \Delta \;=\; B^{2} - 4AC \;=\; 0 - 4 \cdot 4 \cdot (-9) \;=\; 144 \;>\; 0. $$

Conclude.结论。 $\Delta > 0$ $\Rightarrow$ hyperbola. To confirm, complete the square: $4(x^{2} + 4x + 4) - 9(y^{2} - 2y + 1) = 29 + 16 - 9 = 36$, giving $\tfrac{(x+2)^{2}}{9} - \tfrac{(y-1)^{2}}{4} = 1$ , a horizontal hyperbola centred at $(-2, 1)$.$\Delta > 0$ $\Rightarrow$ 双曲线。配方验证:$4(x^{2} + 4x + 4) - 9(y^{2} - 2y + 1) = 29 + 16 - 9 = 36$,得 $\tfrac{(x+2)^{2}}{9} - \tfrac{(y-1)^{2}}{4} = 1$ , 中心在 $(-2, 1)$ 的水平双曲线。

Classify the conic $9 x^{2} + 4 y^{2} - 36 x + 8 y + 4 = 0$.辨识曲线 $9 x^{2} + 4 y^{2} - 36 x + 8 y + 4 = 0$。
§5 · Q1
Ellipse椭圆
Parabola抛物线
Hyperbola双曲线
Circle
$A = 9$, $B = 0$, $C = 4$. $\Delta = 0 - 4 \cdot 9 \cdot 4 = -144 < 0$ $\Rightarrow$ ellipse. Not a circle because $A \ne C$.$A = 9$、$B = 0$、$C = 4$。$\Delta = -144 < 0$ $\Rightarrow$ 椭圆。$A \ne C$ 故不是圆。
Read $A$ and $C$ from the squared coefficients, then $\Delta = -4AC$ when $B = 0$. Negative $\Rightarrow$ ellipse (or circle if $A = C$).由平方项系数读出 $A$、$C$;$B = 0$ 时 $\Delta = -4AC$。为负 $\Rightarrow$ 椭圆($A = C$ 时为圆)。
Classify $3 x^{2} + 4 x y + 2 y^{2} - 5 = 0$.辨识 $3 x^{2} + 4 x y + 2 y^{2} - 5 = 0$。
§5 · Q2
Ellipse (rotated)椭圆(旋转后)
Parabola (rotated)抛物线(旋转后)
Hyperbola (rotated)双曲线(旋转后)
Degenerate / no curve退化 / 无曲线
$A = 3$, $B = 4$, $C = 2$. $\Delta = B^{2} - 4AC = 16 - 24 = -8 < 0$ $\Rightarrow$ ellipse. The presence of $B \ne 0$ means the ellipse is rotated relative to the axes, but the type is still ellipse.$A = 3$、$B = 4$、$C = 2$。$\Delta = B^{2} - 4AC = 16 - 24 = -8 < 0$ $\Rightarrow$ 椭圆。$B \ne 0$ 表示椭圆相对坐标轴旋转,但类型仍为椭圆。
$B^{2} - 4AC = 16 - 24 = -8 < 0$. Negative $\Rightarrow$ ellipse, not parabola. The $xy$ term rotates the curve but does not change which type it is.$B^{2} - 4AC = -8 < 0$。为负 $\Rightarrow$ 椭圆,非抛物线。$xy$ 项使曲线旋转,但不改变类型。

Translations: How $(x - h)$ and $(y - k)$ Move the Centre平移:$(x - h)$ 与 $(y - k)$ 如何移动中心

The translation rule.平移规则。

Replacing $x$ with $(x - h)$ shifts the curve $h$ units right (so the centre moves from $(0, 0)$ to $(h, 0)$). Replacing $y$ with $(y - k)$ shifts it $k$ units up. The same rule applies to every conic, exactly as it did in Unit 10 for general function transformations.把 $x$ 换为 $(x - h)$ 使曲线向右平移 $h$ 单位(中心由 $(0, 0)$ 移至 $(h, 0)$)。把 $y$ 换为 $(y - k)$ 使其向上平移 $k$ 单位。所有圆锥曲线遵循同一规则,与 Unit 10 中函数变换的处理完全一致。

Conic曲线Translated standard form平移后标准式Centre / vertex moves中心 / 顶点移至
Circle$(x - h)^{2} + (y - k)^{2} = r^{2}$$(h, k)$
Parabola (vertical)抛物线(竖轴)$y - k = \tfrac{1}{4p}(x - h)^{2}$vertex $(h, k)$, focus $(h, k + p)$, directrix $y = k - p$顶点 $(h, k)$、焦点 $(h, k + p)$、准线 $y = k - p$
Ellipse椭圆$\tfrac{(x - h)^{2}}{a^{2}} + \tfrac{(y - k)^{2}}{b^{2}} = 1$centre $(h, k)$, foci $(h \pm c, k)$ when $a > b$中心 $(h, k)$;$a > b$ 时焦点 $(h \pm c, k)$
Hyperbola双曲线$\tfrac{(x - h)^{2}}{a^{2}} - \tfrac{(y - k)^{2}}{b^{2}} = 1$centre $(h, k)$, vertices $(h \pm a, k)$, asymptotes $y - k = \pm \tfrac{b}{a}(x - h)$中心 $(h, k)$、顶点 $(h \pm a, k)$、渐近线 $y - k = \pm \tfrac{b}{a}(x - h)$
How to translate from a general form. Complete the square in $x$ and $y$ separately. The constants you add inside the squares are absorbed onto the right-hand side; the leftover gives the centre / vertex offsets. We did this for a circle in §1 and for a hyperbola in §5 , same move for ellipses.由一般式获取平移形式。 分别对 $x$、$y$ 配方。配进平方内的常数被吸收到右端,余下部分给出中心 / 顶点偏移。§1 对圆、§5 对双曲线已做过;椭圆同理。
Worked Example 6 · Translated ellipse from general form例题 6 · 由一般式求平移后的椭圆

Find the centre, vertices, and foci of $4 x^{2} + 9 y^{2} - 16 x + 18 y - 11 = 0$.求 $4 x^{2} + 9 y^{2} - 16 x + 18 y - 11 = 0$ 的中心、顶点与焦点。

Group and factor.分组并提取系数。

$$ 4(x^{2} - 4x) + 9(y^{2} + 2y) = 11. $$

Complete the square inside each group.每组内部配方。

$$ 4(x^{2} - 4x + 4) + 9(y^{2} + 2y + 1) = 11 + 4 \cdot 4 + 9 \cdot 1 = 36. $$ $$ 4(x - 2)^{2} + 9(y + 1)^{2} = 36 \quad\Longrightarrow\quad \frac{(x - 2)^{2}}{9} + \frac{(y + 1)^{2}}{4} = 1. $$

Read off the parts.读出各要素。 Centre $(2, -1)$. $a^{2} = 9 \Rightarrow a = 3$ (horizontal major axis). $b^{2} = 4 \Rightarrow b = 2$. $c = \sqrt{a^{2} - b^{2}} = \sqrt{5}$. Vertices $(2 \pm 3, -1) = (5, -1)$ and $(-1, -1)$. Foci $(2 \pm \sqrt{5}, -1) \approx (4.24, -1)$ and $(-0.24, -1)$.中心 $(2, -1)$。$a^{2} = 9 \Rightarrow a = 3$(主轴水平)。$b^{2} = 4 \Rightarrow b = 2$。$c = \sqrt{a^{2} - b^{2}} = \sqrt{5}$。顶点 $(2 \pm 3, -1) = (5, -1)$、$(-1, -1)$。焦点 $(2 \pm \sqrt{5}, -1) \approx (4.24, -1)$、$(-0.24, -1)$。

Find the centre of the conic $\tfrac{(x + 3)^{2}}{16} - \tfrac{(y - 2)^{2}}{9} = 1$.求曲线 $\tfrac{(x + 3)^{2}}{16} - \tfrac{(y - 2)^{2}}{9} = 1$ 的中心。
§6 · Q1
$(3, -2)$
$(-2, 3)$
$(-3, 2)$
$(3, 2)$
$(x - h)^{2}$ with $h = -3$ matches $(x + 3)^{2}$; $(y - k)^{2}$ with $k = 2$ matches $(y - 2)^{2}$. Centre $(h, k) = (-3, 2)$.$(x - h)^{2}$ 中 $h = -3$ 对应 $(x + 3)^{2}$;$(y - k)^{2}$ 中 $k = 2$ 对应 $(y - 2)^{2}$。中心 $(h, k) = (-3, 2)$。
Match $(x - h)$ to the inside of the squared $x$ term, $(y - k)$ to the inside of the squared $y$ term. Note: $(x + 3) = (x - (-3))$, so $h = -3$.把 $(x - h)$ 对应 $x$ 平方项内部,$(y - k)$ 对应 $y$ 平方项内部。注意 $(x + 3) = (x - (-3))$,故 $h = -3$。
The parabola $y = (x - 5)^{2} - 7$ has vertex at $(5, -7)$. Find its focus and directrix.抛物线 $y = (x - 5)^{2} - 7$ 顶点 $(5, -7)$。求焦点与准线。
§6 · Q2
Focus $(5, -7 + 1)$, directrix $y = -8$焦点 $(5, -7 + 1)$、准线 $y = -8$
Focus $(5, -6.75)$, directrix $y = -7.25$焦点 $(5, -6.75)$、准线 $y = -7.25$
Focus $(5, -7.25)$, directrix $y = -6.75$焦点 $(5, -7.25)$、准线 $y = -6.75$
Focus $(5.25, -7)$, directrix $x = 4.75$焦点 $(5.25, -7)$、准线 $x = 4.75$
Coefficient $1 = \tfrac{1}{4p}$, so $p = 1/4 = 0.25$. Vertex $(5, -7)$, opens up. Focus $(5, -7 + 0.25) = (5, -6.75)$. Directrix $y = -7 - 0.25 = -7.25$.系数 $1 = \tfrac{1}{4p}$,故 $p = 1/4 = 0.25$。顶点 $(5, -7)$,开口向上。焦点 $(5, -7 + 0.25) = (5, -6.75)$。准线 $y = -7 - 0.25 = -7.25$。
From $y - k = \tfrac{1}{4p}(x - h)^{2}$ with coefficient $1$, get $p = 1/4$. Focus is $(h, k + p)$, directrix $y = k - p$.由 $y - k = \tfrac{1}{4p}(x - h)^{2}$ 中系数 $1$ 得 $p = 1/4$。焦点 $(h, k + p)$、准线 $y = k - p$。

Applications: Mirrors, Orbits, Whisper Galleries, Navigation Honors — ON / BC / AB应用:反射镜、轨道、回音廊、导航 荣誉 — ON / BC / AB

Curriculum note.课纲提示。 CCSSM, Ontario, BC, and Alberta do not enumerate parabolic-dish / orbit / whisper-gallery / hyperbolic-navigation applications at standard level. These applications come from the AP Pre-Calc Course and Exam Description (Unit 4) and from IB Math AA HL applications. For Canadian students aiming at engineering / physics streams, §7 is the bridge to first-year university orbital mechanics.CCSSM、安、BC、阿在标准层均未列出抛物面 / 轨道 / 回音廊 / 双曲线导航等应用。这些应用源自 AP Pre-Calc 课纲(Unit 4)与 IB Math AA HL 应用部分。对走工程 / 物理方向的加拿大学生,§7 是通往大学一年级轨道力学的桥梁。
P
Parabolic dishes and mirrors抛物面反射镜与碟形天线

The reflection property of a parabola: any ray entering parallel to the axis reflects through the focus, and any ray leaving the focus reflects out parallel to the axis. This is why parabolic satellite dishes concentrate incoming signals at a receiver placed at the focus, and why car headlights at the focus emit a parallel beam outward. The depth of the dish, the diameter of the rim, and the focus location are all linked by the standard-form parameter $4p$.抛物线的反射性质:所有平行于轴入射的光线都反射汇聚到焦点;所有从焦点出射的光线都反射为平行于轴的光束。这就是为什么抛物面卫星天线把入射信号汇聚到位于焦点的接收器,为什么车头灯在焦点放置光源就能投射平行光束。碟深、口径与焦点位置三者由标准式参数 $4p$ 相连。

Worked example.例题。 A satellite dish has a rim diameter of $2$ m and the dish is $0.25$ m deep at the centre. With the vertex at the origin and axis along $y$, the rim points $(1, 0.25)$ lie on $y = \tfrac{1}{4p} x^{2}$. So $0.25 = \tfrac{1}{4p}$, giving $4p = 4$, $p = 1$. The receiver goes $1$ m above the vertex , right at the rim plane.抛物面天线口径 $2$ m,中心处深 $0.25$ m。顶点置原点、轴沿 $y$,则口缘点 $(1, 0.25)$ 在 $y = \tfrac{1}{4p} x^{2}$ 上。$0.25 = \tfrac{1}{4p}$,$4p = 4$,$p = 1$。接收器应置顶点上方 $1$ m 处 , 恰在口缘平面。

K
Planetary orbits (Kepler's first law)行星轨道(开普勒第一定律)

Kepler's first law (1609): every planet orbits the Sun in an ellipse with the Sun at one focus. Earth's orbit has eccentricity $e \approx 0.0167$ , very close to a circle. Mercury's is $e \approx 0.206$ (visibly elliptical), and Halley's Comet has $e \approx 0.967$ (very elongated). The semi-major axis $a$ is the average orbital radius, and perihelion / aphelion distances are $a(1 - e)$ and $a(1 + e)$ respectively.开普勒第一定律(1609):行星绕日运行的轨道是椭圆,太阳位于其一个焦点。地球轨道离心率 $e \approx 0.0167$,几近圆形;水星 $e \approx 0.206$(明显呈椭圆);哈雷彗星 $e \approx 0.967$(极扁长)。半主轴 $a$ 为平均轨道半径,近日点距离 $a(1 - e)$,远日点距离 $a(1 + e)$。

Worked example.例题。 A comet's orbit has semi-major axis $a = 18$ AU and eccentricity $e = 0.85$. Perihelion: $18 (1 - 0.85) = 2.7$ AU. Aphelion: $18 (1 + 0.85) = 33.3$ AU. So the comet swings from inside Mars's orbit out to past Neptune. The Sun sits at one focus, $c = ea = 15.3$ AU from the centre.某彗星轨道半主轴 $a = 18$ AU,离心率 $e = 0.85$。近日点 $18 (1 - 0.85) = 2.7$ AU;远日点 $18 (1 + 0.85) = 33.3$ AU。彗星从火星轨道内一直摆动到海王星轨道外。太阳位于其一焦点,距椭圆中心 $c = ea = 15.3$ AU。

W
Whisper galleries回音廊

An elliptical room has the property that a sound emitted at one focus reflects off the curved walls and converges at the other focus , even from across the room. Famous examples: the Statuary Hall in the US Capitol; St Paul's Cathedral whispering gallery. The reflection principle is the same as for the parabolic dish, applied to an ellipse: the angles to the two foci are equal at every point of the curve.椭圆形房间具有如下性质:在一焦点发声,经弯曲墙面反射,会汇聚到另一焦点 , 即便两点相隔甚远。著名例子:美国国会大厦雕像厅、伦敦圣保罗大教堂回音廊。其反射原理与抛物面碟相同,应用到椭圆:曲线上任一点处到两焦点的角度相等。

L
Hyperbolic navigation (LORAN)双曲线导航(LORAN)

LORAN (Long Range Navigation), used by ships and aircraft before GPS, exploits the hyperbolic locus directly. Two radio stations transmit synchronised signals; a receiver measures the time difference between the two arrivals. A fixed time difference corresponds to a fixed difference in distance , that is, the receiver lies on a hyperbola with the two stations as foci. A second pair of stations gives a second hyperbola, and the intersection pins down the position. Modern GPS uses a similar principle in 3-D (spheres / hyperboloids).LORAN(远程导航系统),在 GPS 普及前由船舶与飞机使用,正是直接利用双曲线轨迹。两个无线电台同步发射信号;接收器测量两个信号到达的时间差。固定时差对应固定距离差 , 即接收器位于以两电台为焦点的双曲线上。第二组电台给出第二条双曲线,交点即定位。现代 GPS 在三维中沿用类似原理(球面 / 双曲面)。


Exam Strategy and Common Pitfalls考试策略与常见陷阱

Setup discipline解题前的纪律
  • Standard form first, classify second.先化标准式,再分类。 Almost every conic problem starts with "complete the square in $x$ and $y$". Get to the form with $(x - h)^{2}$ and $(y - k)^{2}$ before you read off centres, vertices, or foci.几乎所有圆锥曲线题都从"对 $x$、$y$ 配方"开始。先得到含 $(x - h)^{2}$、$(y - k)^{2}$ 的式子,再读中心、顶点、焦点。
  • Sign of the $x^{2}$ vs $y^{2}$ term.$x^{2}$、$y^{2}$ 符号。 Same sign $\Rightarrow$ ellipse / circle. Opposite signs $\Rightarrow$ hyperbola. Only one squared $\Rightarrow$ parabola. The discriminant $B^{2} - 4AC$ confirms the type when in doubt.同号 $\Rightarrow$ 椭圆 / 圆;异号 $\Rightarrow$ 双曲线;只有一项平方 $\Rightarrow$ 抛物线。存疑时用判别式 $B^{2} - 4AC$ 确认。
  • Ellipse vs hyperbola $c$.椭圆与双曲线的 $c$。 Ellipse: $c^{2} = a^{2} - b^{2}$ (subtraction). Hyperbola: $c^{2} = a^{2} + b^{2}$ (addition). Reversing this is the most common slip on AP / IB problems.椭圆 $c^{2} = a^{2} - b^{2}$(减);双曲线 $c^{2} = a^{2} + b^{2}$(加)。混淆是 AP / IB 题最常见错误。
Circle and parabola tactics (§1-§2, §6)圆与抛物线策略(§1-§2、§6)
  • Circle from three points.三点定圆。 Substitute each into $(x - h)^{2} + (y - k)^{2} = r^{2}$, expand, and subtract pairs to eliminate $h^{2} + k^{2} + r^{2}$ , you get two linear equations in $h, k$. Solve, then back-substitute for $r^{2}$.将三点代入 $(x - h)^{2} + (y - k)^{2} = r^{2}$ 展开后两两相减消去 $h^{2} + k^{2} + r^{2}$,得 $h, k$ 的两个线性方程。解后回代求 $r^{2}$。
  • Parabola sign of $p$.抛物线 $p$ 的正负。 $p > 0$: opens up (vertical) or right (horizontal). $p < 0$: opens down or left. The sign of $\tfrac{1}{4p}$ in vertex form gives this directly , positive coefficient $\Rightarrow$ opens toward the positive axis.$p > 0$:竖轴开口向上 / 横轴开口向右;$p < 0$:开口向下 / 向左。顶点式中 $\tfrac{1}{4p}$ 的符号直接给出 , 系数正 $\Rightarrow$ 开口朝正方向。
Ellipse / Hyperbola / Discriminant tactics (§3-§5) Honors — ON / BC / AB椭圆 / 双曲线 / 判别式策略(§3-§5)荣誉 — ON / BC / AB
  • Identify the major axis direction first.先判断主轴方向。 For an ellipse, $a$ is the larger semi-axis; whichever variable has the larger squared denominator owns the major axis. For a hyperbola, the positive term owns the transverse axis (no "larger" rule).椭圆中 $a$ 为较长半轴;分母平方较大者所对应的变量方向即为主轴。双曲线中正项所在的变量方向即为实轴(没有"较大者"规则)。
  • Asymptote sketch from the rectangle.由渐近矩形作渐近线。 For a hyperbola $\tfrac{(x-h)^{2}}{a^{2}} - \tfrac{(y-k)^{2}}{b^{2}} = 1$, draw the rectangle with corners $(h \pm a, k \pm b)$; the asymptotes are the lines through opposite corners.对双曲线 $\tfrac{(x-h)^{2}}{a^{2}} - \tfrac{(y-k)^{2}}{b^{2}} = 1$,作以 $(h \pm a, k \pm b)$ 为四角的矩形;渐近线为过两对角的两条直线。
  • Eccentricity as the unifying parameter.离心率作为统一参数。 $e = 0$ circle; $0 < e < 1$ ellipse; $e = 1$ parabola (defined separately); $e > 1$ hyperbola. AP Pre-Calc Unit 3 explicitly uses this scale.$e = 0$ 圆;$0 < e < 1$ 椭圆;$e = 1$ 抛物线(单独定义);$e > 1$ 双曲线。AP Pre-Calc Unit 3 明确采用此刻度。
  • Discriminant before completing the square.先用判别式,再配方。 Compute $B^{2} - 4AC$ in one line to know which type you're chasing, then commit to the (rotation +) translation needed for the standard form.先一行算出 $B^{2} - 4AC$ 知道类型,再按需进行(旋转 +)平移化标准式。
Applied-problem hygiene应用题规范
  • Place coordinates carefully.慎选坐标系。 Parabolic dish: vertex at origin, axis along $y$, rim symmetric about $x = 0$. Elliptical orbit: centre of ellipse at origin, foci on $x$-axis. The right coordinate choice makes the algebra short.抛物面碟:顶点在原点,轴沿 $y$,口缘关于 $x = 0$ 对称。椭圆轨道:椭圆中心在原点,焦点在 $x$ 轴。坐标选对,代数即短。
  • Units in Kepler problems.开普勒题中的单位。 Astronomical units (AU) for distance from Sun; years for period. Kepler's third law $T^{2} = a^{3}$ holds with these specific units , not metres / seconds.距日距离用天文单位(AU);周期用年。开普勒第三定律 $T^{2} = a^{3}$ 仅在这套单位下成立,不能换为米 / 秒。
  • Answer in a sentence with units.用整句加单位作答。 "The receiver sits $1$ m above the vertex." earns the communication mark; "$1$" alone may not."接收器置于顶点上方 $1$ m" 可拿到表达分;只写 "$1$" 可能拿不到。

Flashcards闪卡

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Distance formula?距离公式?
$$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$$
Circle equation (centre $(h, k)$, radius $r$)?圆方程(圆心 $(h, k)$、半径 $r$)?
$$(x - h)^{2} + (y - k)^{2} = r^{2}$$ CCSSM HSG-GPE.A.1CCSSM HSG-GPE.A.1
Parabola locus definition?抛物线轨迹定义?
Equidistant from focus and directrix.到焦点与准线等距。 $$\mathrm{dist}(P, F) = \mathrm{dist}(P, \text{directrix})$$
Vertical-axis parabola standard form?竖轴抛物线标准式?
$$y - k = \tfrac{1}{4p}(x - h)^{2}$$ vertex $(h, k)$, focus $(h, k+p)$, directrix $y = k-p$顶点 $(h, k)$、焦点 $(h, k+p)$、准线 $y = k-p$
Ellipse locus definition?椭圆轨迹定义?
$$|PF_{1}| + |PF_{2}| = 2a$$
Ellipse standard form ($a > b > 0$)?椭圆标准式($a > b > 0$)?
$$\tfrac{x^{2}}{a^{2}} + \tfrac{y^{2}}{b^{2}} = 1$$ $c^{2} = a^{2} - b^{2}$$c^{2} = a^{2} - b^{2}$
Hyperbola locus definition?双曲线轨迹定义?
$$\big| |PF_{1}| - |PF_{2}| \big| = 2a$$
Hyperbola standard form (horizontal)?双曲线标准式(水平实轴)?
$$\tfrac{x^{2}}{a^{2}} - \tfrac{y^{2}}{b^{2}} = 1$$ $c^{2} = a^{2} + b^{2}$, asymptotes $y = \pm \tfrac{b}{a} x$$c^{2} = a^{2} + b^{2}$,渐近线 $y = \pm \tfrac{b}{a} x$
Eccentricity?离心率?
$$e = \tfrac{c}{a}$$ $e = 0$ circle; $0 < e < 1$ ellipse; $e = 1$ parabola; $e > 1$ hyperbola$e = 0$ 圆;$0 < e < 1$ 椭圆;$e = 1$ 抛物线;$e > 1$ 双曲线
Conic discriminant?圆锥曲线判别式?
$$\Delta = B^{2} - 4AC$$ $< 0$ ellipse; $= 0$ parabola; $> 0$ hyperbola$< 0$ 椭圆;$= 0$ 抛物线;$> 0$ 双曲线
Sign of $c^{2}$: ellipse vs hyperbola?椭圆与双曲线 $c^{2}$ 符号?
Ellipse $c^{2} = a^{2} - b^{2}$. Hyperbola $c^{2} = a^{2} + b^{2}$.椭圆 $c^{2} = a^{2} - b^{2}$;双曲线 $c^{2} = a^{2} + b^{2}$。
Translate a conic to centre $(h, k)$?把圆锥曲线平移到中心 $(h, k)$?
Replace $x \to (x - h)$ and $y \to (y - k)$ in the standard form.在标准式中把 $x$ 换为 $(x - h)$,$y$ 换为 $(y - k)$。
Parabolic-dish receiver location?抛物面碟接收器位置?
At the focus. Parallel rays reflect through the focus.置于焦点。平行入射光线反射汇聚到焦点。
Kepler perihelion / aphelion?开普勒近日 / 远日点?
$$r_{\min} = a(1 - e), \;\; r_{\max} = a(1 + e)$$

Practice Quiz综合测验

Find the equation of the circle through $(0, 0)$, $(6, 0)$, $(0, 8)$.求过 $(0, 0)$、$(6, 0)$、$(0, 8)$ 的圆方程。
Q1
$(x - 3)^{2} + (y - 4)^{2} = 5$
$x^{2} + y^{2} = 100$
$(x - 3)^{2} + (y - 4)^{2} = 25$
$(x - 6)^{2} + (y - 8)^{2} = 25$
The three points form a right triangle with the right angle at $(0, 0)$, so the hypotenuse from $(6, 0)$ to $(0, 8)$ is a diameter. Its midpoint $(3, 4)$ is the centre. Diameter length $\sqrt{36 + 64} = 10$, so $r = 5$ and $r^{2} = 25$.三点构成在 $(0, 0)$ 处有直角的直角三角形,故 $(6, 0)$ 到 $(0, 8)$ 的斜边即为直径。中点 $(3, 4)$ 即圆心。直径长 $\sqrt{36 + 64} = 10$,故 $r = 5$、$r^{2} = 25$。
Spot the right angle at $(0, 0)$ , the diameter is the opposite hypotenuse, and the centre is its midpoint.看出 $(0, 0)$ 处为直角 , 对面斜边即直径,圆心为其中点。
A parabola with vertex $(0, 0)$ has directrix $x = -3$. Find its equation.顶点 $(0, 0)$、准线 $x = -3$ 的抛物线方程。
Q2
$y^{2} = 12 x$
$x^{2} = 12 y$
$y^{2} = -12 x$
$x^{2} = -12 y$
Directrix vertical $\Rightarrow$ horizontal axis. Directrix to left of vertex $\Rightarrow$ opens right, $p > 0$. $p = 3$, so $x = \tfrac{1}{4p} y^{2} = \tfrac{1}{12} y^{2}$, i.e. $y^{2} = 12 x$.准线竖直 $\Rightarrow$ 横轴。准线在顶点左侧 $\Rightarrow$ 开口向右,$p > 0$。$p = 3$,故 $x = \tfrac{1}{12} y^{2}$,即 $y^{2} = 12 x$。
Vertical directrix means horizontal axis. Use $x = \tfrac{1}{4p} y^{2}$ with $p$ = signed distance from vertex to focus (here $+3$).准线竖直 $\Rightarrow$ 横轴。用 $x = \tfrac{1}{4p} y^{2}$,$p$ 为顶点到焦点的有向距离(此处 $+3$)。
An ellipse has foci $(\pm 3, 0)$ and passes through $(5, 0)$. Find its standard equation. 🇺🇸 HSG-GPE.A.3 (+)椭圆焦点 $(\pm 3, 0)$ 且过 $(5, 0)$。求标准方程。🇺🇸 HSG-GPE.A.3 (+)
Q3
$\tfrac{x^{2}}{25} + \tfrac{y^{2}}{9} = 1$
$\tfrac{x^{2}}{25} + \tfrac{y^{2}}{16} = 1$
$\tfrac{x^{2}}{16} + \tfrac{y^{2}}{25} = 1$
$\tfrac{x^{2}}{34} + \tfrac{y^{2}}{25} = 1$
$(5, 0)$ is a vertex on the major axis, so $a = 5$, $a^{2} = 25$. $c = 3$, so $b^{2} = a^{2} - c^{2} = 25 - 9 = 16$. Equation $\tfrac{x^{2}}{25} + \tfrac{y^{2}}{16} = 1$.$(5, 0)$ 是主轴顶点,$a = 5$,$a^{2} = 25$。$c = 3$,$b^{2} = a^{2} - c^{2} = 16$。方程 $\tfrac{x^{2}}{25} + \tfrac{y^{2}}{16} = 1$。
For an ellipse, $b^{2} = a^{2} - c^{2}$. The given point's distance from centre along the major axis is $a$.椭圆 $b^{2} = a^{2} - c^{2}$。给定点沿主轴方向到中心的距离即 $a$。
For $\tfrac{x^{2}}{16} - \tfrac{y^{2}}{20} = 1$, find $c$. 🇺🇸 HSG-GPE.A.3 (+)对 $\tfrac{x^{2}}{16} - \tfrac{y^{2}}{20} = 1$,求 $c$。🇺🇸 HSG-GPE.A.3 (+)
Q4
$2$
$4$
$6$
$\sqrt{20}$
Hyperbola: $c^{2} = a^{2} + b^{2} = 16 + 20 = 36$, so $c = 6$.双曲线:$c^{2} = a^{2} + b^{2} = 16 + 20 = 36$,故 $c = 6$。
For a hyperbola the relation is $c^{2} = a^{2} + b^{2}$ , addition. The ellipse uses subtraction.双曲线为 $c^{2} = a^{2} + b^{2}$(加),椭圆才是减。
Classify $2 x^{2} - 5 y^{2} + 3 x - 7 = 0$. 🇺🇸 AP Pre-Calc辨识 $2 x^{2} - 5 y^{2} + 3 x - 7 = 0$。🇺🇸 AP Pre-Calc
Q5
Ellipse椭圆
Hyperbola双曲线
Parabola抛物线
Circle
$A = 2$, $B = 0$, $C = -5$. $\Delta = 0 - 4 \cdot 2 \cdot (-5) = 40 > 0$ $\Rightarrow$ hyperbola. (Quicker: $x^{2}$ and $y^{2}$ have opposite signs, so it's a hyperbola.)$A = 2$、$B = 0$、$C = -5$。$\Delta = 40 > 0$ $\Rightarrow$ 双曲线。(更快:$x^{2}$、$y^{2}$ 异号 $\Rightarrow$ 双曲线。)
Opposite signs on $x^{2}$ and $y^{2}$ already tell you "hyperbola". The discriminant $B^{2} - 4AC = -4AC$ confirms it when $B = 0$.$x^{2}$、$y^{2}$ 异号即可断为双曲线。$B = 0$ 时判别式 $B^{2} - 4AC = -4AC$ 给出确认。
Find the centre and radius of $x^{2} + y^{2} - 10 x + 4 y + 13 = 0$.求 $x^{2} + y^{2} - 10 x + 4 y + 13 = 0$ 的圆心与半径。
Q6
Centre $(10, -4)$, radius $13$圆心 $(10, -4)$、半径 $13$
Centre $(-5, 2)$, radius $4$圆心 $(-5, 2)$、半径 $4$
Centre $(5, -2)$, radius $4$圆心 $(5, -2)$、半径 $4$
Centre $(5, -2)$, radius $\sqrt{42}$圆心 $(5, -2)$、半径 $\sqrt{42}$
$(x^{2} - 10x + 25) + (y^{2} + 4y + 4) = -13 + 25 + 4 = 16$, giving $(x - 5)^{2} + (y + 2)^{2} = 16$. Centre $(5, -2)$, radius $4$.$(x^{2} - 10x + 25) + (y^{2} + 4y + 4) = -13 + 25 + 4 = 16$,得 $(x - 5)^{2} + (y + 2)^{2} = 16$。圆心 $(5, -2)$、半径 $4$。
Complete the square in $x$ and $y$, then the right-hand-side equals $r^{2}$. Sign flips: $x^{2} - 10x$ goes to $(x - 5)^{2}$, not $(x + 5)^{2}$.分别配方后右端等于 $r^{2}$。注意符号:$x^{2} - 10x$ 化为 $(x - 5)^{2}$,不是 $(x + 5)^{2}$。
A parabolic dish has rim diameter $4$ m and depth $0.5$ m. Where is the focus? 🇺🇸 AP Pre-Calc applications抛物面碟口径 $4$ m,深 $0.5$ m。焦点在何处?🇺🇸 AP Pre-Calc 应用
Q7
$2$ m above the vertex顶点上方 $2$ m
$0.5$ m above the vertex顶点上方 $0.5$ m
$4$ m above the vertex顶点上方 $4$ m
$1$ m above the vertex顶点上方 $1$ m
Vertex at origin, axis along $y$. Rim point $(2, 0.5)$ on $y = \tfrac{1}{4p} x^{2}$: $0.5 = \tfrac{1}{4p}(4) = \tfrac{1}{p}$, so $p = 2$. Focus is at $(0, 2)$ , $2$ m above the vertex.顶点在原点、轴沿 $y$。口缘点 $(2, 0.5)$ 在 $y = \tfrac{1}{4p} x^{2}$ 上:$0.5 = \tfrac{1}{4p}(4) = \tfrac{1}{p}$,故 $p = 2$。焦点 $(0, 2)$ , 顶点上方 $2$ m。
Place vertex at origin and substitute the rim point into $y = \tfrac{1}{4p} x^{2}$. Solve for $p$; focus sits at $(0, p)$.顶点置原点,把口缘点代入 $y = \tfrac{1}{4p} x^{2}$ 解出 $p$;焦点在 $(0, p)$。
Halley's Comet has semi-major axis $a \approx 17.8$ AU and eccentricity $e \approx 0.967$. Find its perihelion distance. 🇺🇸 AP Pre-Calc applications哈雷彗星 $a \approx 17.8$ AU、$e \approx 0.967$。求近日点距离。🇺🇸 AP Pre-Calc 应用
Q8
$\approx 35.0$ AU
$\approx 0.59$ AU
$\approx 17.2$ AU
$\approx 1.83$ AU
$r_{\min} = a(1 - e) = 17.8 \cdot (1 - 0.967) = 17.8 \cdot 0.033 \approx 0.59$ AU. (Aphelion $r_{\max} = a(1 + e) \approx 35.0$ AU is choice (a).)$r_{\min} = a(1 - e) = 17.8 \cdot 0.033 \approx 0.59$ AU。(远日点 $r_{\max} = a(1 + e) \approx 35.0$ AU 即选项 (a)。)
Perihelion $= a(1 - e)$ , closest distance. Aphelion $= a(1 + e)$ , farthest.近日点 $= a(1 - e)$(最近);远日点 $= a(1 + e)$(最远)。
Find the asymptotes of $\tfrac{(x - 1)^{2}}{4} - \tfrac{(y + 2)^{2}}{9} = 1$. 🇺🇸 HSG-GPE.A.3 (+)求 $\tfrac{(x - 1)^{2}}{4} - \tfrac{(y + 2)^{2}}{9} = 1$ 的渐近线。🇺🇸 HSG-GPE.A.3 (+)
Q9
$y + 2 = \pm \tfrac{3}{2}(x - 1)$
$y - 2 = \pm \tfrac{3}{2}(x + 1)$
$y + 2 = \pm \tfrac{2}{3}(x - 1)$
$y = \pm \tfrac{3}{2} x$
$a = 2$, $b = 3$, centre $(1, -2)$. Horizontal transverse axis $\Rightarrow$ asymptotes $y - k = \pm \tfrac{b}{a} (x - h)$, i.e. $y + 2 = \pm \tfrac{3}{2}(x - 1)$.$a = 2$、$b = 3$、中心 $(1, -2)$。实轴水平 $\Rightarrow$ 渐近线 $y - k = \pm \tfrac{b}{a} (x - h)$,即 $y + 2 = \pm \tfrac{3}{2}(x - 1)$。
For a translated hyperbola, asymptotes pass through the centre with slopes $\pm b / a$ (horizontal transverse) or $\pm a / b$ (vertical).平移后的双曲线,渐近线过中心,斜率为 $\pm b / a$(水平实轴)或 $\pm a / b$(竖直实轴)。
A whisper gallery is shaped as half an ellipse with $a = 12$ m and $b = 5$ m. How far apart are the two foci?某回音廊截面为半椭圆,$a = 12$ m、$b = 5$ m。两焦点相距多远?
Q10
$13$ m
$7$ m
$\approx 21.8$ m
$24$ m
$c = \sqrt{a^{2} - b^{2}} = \sqrt{144 - 25} = \sqrt{119} \approx 10.91$ m. Two foci are at $(\pm c, 0)$, so the distance between them is $2c \approx 21.8$ m. Whisper from one focus reaches the other after a single reflection , the locus property in action.$c = \sqrt{a^{2} - b^{2}} = \sqrt{119} \approx 10.91$ m。两焦点在 $(\pm c, 0)$,相距 $2c \approx 21.8$ m。在一焦点低语,经单次反射可达另一焦点 , 轨迹性质付诸实用。
Ellipse: $c = \sqrt{a^{2} - b^{2}}$. Distance between foci is $2c$, not $c$.椭圆 $c = \sqrt{a^{2} - b^{2}}$。两焦点距离为 $2c$,不是 $c$。

Readiness Checklist准备就绪清单

Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成,再勾选每一项。

0 / 12 mastered已掌握 0 / 12

What This Feeds Into本单元的去向

Conic sections is the geometric foundation for parametric curves, polar equations, and orbital mechanics. The next HS Math units take the locus-equation thinking forward into vectors (Unit 14), where lines and planes appear with the same focus-directrix flavour, and into limits / continuity (Unit 15). At the AP / IB layer, conic sections reappear inside AP Pre-Calc Unit 4 (parametric and polar curves), inside IB Math AA HL Topic B1 (functions and equations including conics), and inside AP Physics C orbital mechanics.圆锥曲线是参数曲线、极坐标方程与轨道力学的几何基础。HS Math 后续单元把"轨迹-方程"思路带入向量(Unit 14;直线与平面以类似焦点-准线方式出现)以及极限 / 连续性(Unit 15)。AP / IB 层面,圆锥曲线在 AP Pre-Calc Unit 4(参数与极坐标曲线)、IB Math AA HL Topic B1(函数与方程含圆锥曲线)以及 AP Physics C 轨道力学中重现。

Within High School Math.在 HS Math 内部。

Unit 10 (Function Transformations) supplies the $(x - h)$, $(y - k)$ shifting machinery that §6 reuses for every conic. Unit 13 (Probability) is unrelated. Unit 14 (Vectors) parametrises lines and planes; the locus thinking from this unit applies directly. Unit 15 (Limits) uses the asymptotes of §4 as a canonical motivating example of "approaching but never reaching".Unit 10(函数变换)提供 $(x - h)$、$(y - k)$ 平移机制 , §6 在每条曲线上复用。Unit 13(概率)与本单元无关。Unit 14(向量)参数化直线与平面;本单元的轨迹思想直接适用。Unit 15(极限)以 §4 的渐近线作为"无限趋近但不到达"的典型例子。

Across the AP and IB feeders in this repo.本仓库中的 AP 与 IB 衔接单元。

IB Math HL B1 · Functions and Equations (locus thinking and standard forms reappear at HL depth)IB Math HL B1 · 函数与方程(轨迹思想与标准式在 HL 级别重现) IB Math HL C3 · Vectors (parametric lines and planes , same locus-equation flavour as conics)IB Math HL C3 · 向量(参数化直线与平面 , 与圆锥曲线"轨迹-方程"思路同源) AP Calculus Unit 9 · Parametric, Polar, and Vector-Valued Functions (ellipses and hyperbolas in polar form)AP Calculus Unit 9 · 参数方程、极坐标与向量函数(极坐标下的椭圆与双曲线) AP Physics Unit 1 · Kinematics (projectile motion traces a parabola , the §2 standard form in disguise)AP Physics Unit 1 · 运动学(抛体轨迹即抛物线 , 隐藏的 §2 标准式)

If you are aiming for the SAT, the circle equation from §1 is the most-tested conic. If you are aiming for AP Calculus, you need fluent parabola / ellipse / hyperbola standard forms for Unit 9 (parametric and polar curves). For IB Math AA HL, all four conics live in Topic B1, and the discriminant test from §5 is the fastest path through any "what kind of curve is this?" exam question.备考 SAT:§1 的圆方程是测得最多的圆锥曲线题。备考 AP Calculus:Unit 9(参数与极坐标)要求熟练抛物线 / 椭圆 / 双曲线的标准式。备考 IB Math AA HL:四条曲线均在 Topic B1;§5 的判别式是解决"这是哪种曲线"类题最快的路径。