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Conic Sections · Solutions圆锥曲线 · 详解

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EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 🇨🇦 AB阿 SAT-style MCQSAT 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 AB Provincial-style阿尔伯塔省考风格 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解SAT MCQ + ON / AB short answer · 18 marksSAT 选择题 + 安 / 阿省考短答 · 共 18 分

Section A · Worked SolutionsA 部分 · 详细解答

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]

Centre and radius of $(x - 2)^{2} + (y + 5)^{2} = 49$.求 $(x - 2)^{2} + (y + 5)^{2} = 49$ 的圆心与半径。

Answer:答案：  (B)  Centre $(2, -5)$, radius $7$圆心 $(2, -5)$，半径 $7$

(a) Read centre and radius off the standard form从标准式读出圆心与半径 M1·A1·A1

Standard form is $(x - h)^{2} + (y - k)^{2} = r^{2}$. Match: $h = 2$ (the value zeroing the $x$-bracket), $k = -5$ (since $y + 5 = y - (-5)$), and $r^{2} = 49 \Rightarrow r = 7$. Centre $(2, -5)$, radius $7$, option (B).标准式为 $(x - h)^{2} + (y - k)^{2} = r^{2}$。对照：使 $x$ 括号为零的 $h = 2$；因为 $y + 5 = y - (-5)$，所以 $k = -5$；$r^{2} = 49 \Rightarrow r = 7$。圆心 $(2, -5)$，半径 $7$，选 (B)。
Why the wrong choices fail.干扰项分析。
  • (A) $(-2, 5)$, $r = 7$, both centre signs flipped, the classic "read the bracket sign without solving" error.：圆心两坐标均符号反，是典型"不解直接读括号里的符号"错误。
  • (C) $(2, -5)$, $r = 49$, correct centre but forgot to take the square root of $49$ to get $r$.：圆心正确但忘了对 $49$ 开方得 $r$。
  • (D) $(-2, 5)$, $r = 49$, both errors stacked.：两类错叠加。
Sign-flip on the bracket, square-root on the right.括号里要变号，等号右边要开方。 Standard form encodes $h, k$ as the values that zero the brackets, so $(x - 2)$ gives $h = 2$ and $(y + 5) = (y - (-5))$ gives $k = -5$. The right-hand side is $r^{2}$, never $r$. This is the same $(h, k)$-from-bracket move you used for vertex form in Unit 2 (parabola vertex) and Unit 5 (transformations); the only new piece in Unit 12 is the right-hand side carrying $r^{2}$.标准式中 $h, k$ 是让括号等于零的值，故 $(x - 2)$ 给出 $h = 2$，$(y + 5) = (y - (-5))$ 给出 $k = -5$。等号右边是 $r^{2}$，绝不是 $r$。这跟你在第 2 单元（抛物线顶点）和第 5 单元（图像变换）用过的"从括号读 $(h, k)$"是同一招；第 12 单元唯一新增的是右边变成 $r^{2}$。
Q2EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §2 Parabola Focus-Directrix抛物线焦点—准线 · HSG-GPE.A.2 [3 marks][3 分]

Parabola with focus $(0, 3)$, directrix $y = -3$.焦点 $(0, 3)$、准线 $y = -3$ 的抛物线。

Answer:答案：  (B)  $y = \tfrac{1}{12} x^{2}$

(a) Identify vertex and focal length $p$确定顶点与焦距 $p$ M1·A1

Vertex is midway between focus and directrix: midpoint of $(0, 3)$ and the line $y = -3$ is $(0, 0)$, so the parabola has vertex at the origin. The focal length is $p = 3$ (distance from vertex to focus).顶点位于焦点与准线的中点：点 $(0, 3)$ 与直线 $y = -3$ 的中点是 $(0, 0)$，故顶点在原点。焦距 $p = 3$（顶点到焦点的距离）。

(b) Apply the focus-directrix form套用焦点—准线形式 A1

A parabola opening upward with vertex at the origin and focal length $p$ has equation $x^{2} = 4 p y$, i.e. $y = \dfrac{x^{2}}{4 p}$. With $p = 3$: $y = \dfrac{x^{2}}{12}$. Matches (B).顶点在原点、开口向上、焦距为 $p$ 的抛物线方程为 $x^{2} = 4 p y$，即 $y = \dfrac{x^{2}}{4 p}$。代入 $p = 3$：$y = \dfrac{x^{2}}{12}$，对应 (B)。
Why the wrong choices fail.干扰项分析。
  • (A) $y = \tfrac{1}{6} x^{2}$, used $4p = 6$ (forgot $p$ is the half-distance from focus to directrix, not the full distance).：误用 $4p = 6$（忘记 $p$ 是焦点到准线距离的一半，而非全距离）。
  • (C) $y = 6 x^{2}$, inverted the coefficient (wrote $6$ instead of $1/(4 \cdot 3) = 1/12$).：系数取了倒数错位（写成 $6$ 而非 $1/(4 \cdot 3) = 1/12$）。
  • (D) $y = 12 x^{2}$, inverted and used $4p$ as the coefficient instead of $1/(4p)$.：把 $4p$ 当作系数而非 $1/(4p)$。
Focus-directrix gives you $p$ for free, vertex is the midpoint.焦点—准线立即给你 $p$，顶点就是中点。 The focus-directrix definition says every point on the parabola is equidistant from focus and directrix. The vertex is the closest point on the parabola to the directrix, so it sits halfway between focus and directrix; $p$ is that half-distance. Internalising $x^{2} = 4 p y$ as "$\;4 \cdot \text{focal length} \cdot y = x^{2}\;$" prevents the inversion errors in (C) and (D). This is the form AP Pre-Calculus and HSG-GPE.A.2 will lean on every time.焦点—准线定义说抛物线上每点到焦点与准线的距离相等。顶点是抛物线上离准线最近的点，故位于焦点与准线的正中间；$p$ 就是这半段距离。把 $x^{2} = 4 p y$ 记成"$\;4 \cdot \text{焦距} \cdot y = x^{2}\;$"可避免 (C)、(D) 那类取倒数错位。这是 AP Pre-Calculus 与 HSG-GPE.A.2 反复倚重的形式。
Q3MEDIUM中 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §3 Ellipse Eccentricity椭圆离心率 · HSG-GPE.A.3 (+) [3 marks][3 分]

Eccentricity of $\tfrac{x^{2}}{25} + \tfrac{y^{2}}{9} = 1$.求 $\tfrac{x^{2}}{25} + \tfrac{y^{2}}{9} = 1$ 的离心率。

Answer:答案：  (B)  $e = \tfrac{4}{5}$

(a) Identify $a^{2}, b^{2}$, then compute $c^{2} = a^{2} - b^{2}$识别 $a^{2}, b^{2}$，再算 $c^{2} = a^{2} - b^{2}$ M1·A1

Standard form $\dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} = 1$ with $a^{2} = 25 > 9 = b^{2}$, so $a = 5$, $b = 3$ and the major axis is horizontal. Focal relation:标准式 $\dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} = 1$，其中 $a^{2} = 25 > 9 = b^{2}$，故 $a = 5$、$b = 3$，长轴为水平方向。焦点关系： $$ c^{2} \;=\; a^{2} - b^{2} \;=\; 25 - 9 \;=\; 16 \;\Longrightarrow\; c \;=\; 4. $$

(b) Apply $e = c/a$套用 $e = c/a$ A1

Eccentricity $e = \dfrac{c}{a} = \dfrac{4}{5}$. Matches (B).离心率 $e = \dfrac{c}{a} = \dfrac{4}{5}$，对应 (B)。
Why the wrong choices fail.干扰项分析。
  • (A) $e = \tfrac{3}{5}$, used $b/a$ instead of $c/a$, a common slip (the "wrong axis" trap).：误用 $b/a$ 而非 $c/a$，是常见的"用错轴"陷阱。
  • (C) $e = \tfrac{5}{3}$, inverted $c/a$ to $a/c$; this would give $e > 1$, which is impossible for an ellipse (eccentricity of an ellipse always lies in $[0, 1)$).：把 $c/a$ 颠倒为 $a/c$；这会给出 $e > 1$，椭圆不可能（椭圆离心率必在 $[0, 1)$ 内）。
  • (D) $e = \tfrac{16}{25}$, used $c^{2}/a^{2}$ without taking the square root.：用了 $c^{2}/a^{2}$ 但忘了开方。
$e \in [0, 1)$ for an ellipse, $e = 1$ for a parabola, $e > 1$ for a hyperbola.椭圆 $e \in [0, 1)$，抛物线 $e = 1$，双曲线 $e > 1$。 Eccentricity is a single dimensionless number that classifies the conic and measures how "stretched" it is. For an ellipse, $e = 0$ is a circle ($c = 0$, foci coincide at the centre); $e \to 1$ stretches the ellipse toward a parabolic limit. Any answer with $e \ge 1$ on an ellipse question is automatically a distractor, knowing the range filters half the wrong choices instantly.离心率是一个无量纲数，既能分辨圆锥曲线类型，又能衡量它"被拉伸"的程度。椭圆中 $e = 0$ 即圆（$c = 0$，两焦点重合于中心）；$e \to 1$ 把椭圆拉向抛物线极限。任何椭圆题给出 $e \ge 1$ 的选项必为干扰项——记住范围能瞬间筛掉一半错选。
Q4MEDIUM 🇨🇦 ON ON Provincial-style §1 Distance, Midpoint, Circle · MPM2D Analytic Geometry [4 marks]

$A(-1, 2)$, $B(7, 8)$ are endpoints of a diameter. (a) Centre. (b) Radius. (c) Standard-form equation.

Answer:  (a) centre $(3, 5)$  ·  (b) $r = 5$  ·  (c) $(x - 3)^{2} + (y - 5)^{2} = 25$

(a) Midpoint of diameter A1

Centre $=$ midpoint of $AB = \left(\dfrac{-1 + 7}{2}, \dfrac{2 + 8}{2}\right) = (3, 5)$.

(b) Radius via distance from centre to endpoint M1·A1

$r = \text{dist}((3, 5), (-1, 2)) = \sqrt{(3 - (-1))^{2} + (5 - 2)^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5$. (Sanity check: $r$ is also half the diameter $|AB| = \sqrt{64 + 36} = \sqrt{100} = 10$, so $r = 5$. $\checkmark$)

(c) Substitute into standard form A1

$(x - 3)^{2} + (y - 5)^{2} = 5^{2} = 25$.
Two distance-formula uses, two cross-checks. When the diameter endpoints are given, you can compute the radius two independent ways: (i) the full diameter $|AB|$ then halve, (ii) from the centre (midpoint) to either endpoint. Both should agree. ON MPM2D markers explicitly reward students who run the cross-check, because it catches midpoint arithmetic errors that would otherwise propagate. The 3-4-5 Pythagorean triple jumping out of $\sqrt{16 + 9}$ is the kind of "clean answer" that confirms the algebra.
Q5MEDIUM 🇨🇦 AB AB Provincial-style §2 Parabola, Vertex Form · AB Math 20-1 RF GO 3 [5 marks]

Vertex $(2, -3)$, opens upward, $a = 1/4$. (a) Vertex form. (b) $p$. (c) Focus + directrix. (d) AOS.

Answer:  (a) $y = \tfrac{1}{4}(x - 2)^{2} - 3$  ·  (b) $p = 1$  ·  (c) focus $(2, -2)$, directrix $y = -4$  ·  (d) $x = 2$

(a) Substitute $(h, k, a)$ into vertex form A1

With $h = 2$, $k = -3$, $a = \tfrac{1}{4}$: $$ y \;=\; \tfrac{1}{4}(x - 2)^{2} - 3. $$

(b) Solve $4 p = 1/a$ A1

$4 p = \dfrac{1}{a} = \dfrac{1}{1/4} = 4 \Rightarrow p = 1$.

(c) Focus + directrix A1·A1

For an upward-opening parabola with vertex $(h, k)$, focus is $(h, k + p)$ and directrix is $y = k - p$.
Focus: $(2, -3 + 1) = (2, -2)$. Directrix: $y = -3 - 1 = -4$.

(d) Axis of symmetry A1

Axis of symmetry: the vertical line through the vertex, $x = 2$.
$4p = 1/a$ is the bridge between vertex-form quadratic and focus-directrix parabola. AB Math 20-1 introduces $y = a(x - h)^{2} + k$ as a transformed quadratic but does not formalise focus and directrix, those live in the US HSG-GPE.A.2 standard and in AP Pre-Calculus. Memorising the conversion $4p = 1/a$ (equivalently $a = 1/(4p)$) lets you switch between the two notations in one line: large $|a|$ $\Leftrightarrow$ small $p$ $\Leftrightarrow$ "tight" parabola (focus close to vertex). This is the same conversion that powered the dish question in Part III.
PART II  ·  EXTENDED RESPONSE · SOLUTIONSAP-feeder FRQ + honors · 35 marks

Section B · Worked Solutions

Q6MEDIUM 🇺🇸 US AP-feeder FRQ §3 Ellipse Standard Form · HSG-GPE.A.3 (+) [8 marks]

Ellipse: centre origin, major along $x$-axis, $a = 5$, $b = 4$. (a) Equation. (b) Vertices. (c) Foci via $c^{2} = a^{2} - b^{2}$. (d) Eccentricity.

Answer:  (a) $\tfrac{x^{2}}{25} + \tfrac{y^{2}}{16} = 1$  ·  (b) $(\pm 5, 0)$, $(0, \pm 4)$  ·  (c) $c = 3$, foci $(\pm 3, 0)$  ·  (d) $e = 3/5 = 0.6$, modestly elongated

(a) Standard form A1

$\dfrac{x^{2}}{5^{2}} + \dfrac{y^{2}}{4^{2}} = \dfrac{x^{2}}{25} + \dfrac{y^{2}}{16} = 1$.

(b) Four vertices A1·A1

Endpoints of the major axis (along $x$): $(\pm a, 0) = (\pm 5, 0)$.
Endpoints of the minor axis (along $y$): $(0, \pm b) = (0, \pm 4)$.

(c) Foci via focal relation M1·A1·A1

$c^{2} = a^{2} - b^{2} = 25 - 16 = 9 \Rightarrow c = 3$. Foci lie on the major axis (the longer one), so $F_{1, 2} = (\pm 3, 0)$.

(d) Eccentricity and classification A1·A1

$e = \dfrac{c}{a} = \dfrac{3}{5} = 0.6$. Since $0 < e < 1$ and $e$ is closer to $0$ than to $1$, the ellipse is modestly elongated, noticeably non-circular but not extreme. (Earth's orbit, by contrast, has $e \approx 0.017$, "nearly circular".)
Foci go on the major axis, always. The focal relation $c^{2} = a^{2} - b^{2}$ requires $a > b$ by convention: $a$ is always the semi-major axis. The two foci sit on the major axis equidistant from the centre, never on the minor axis. AP markers explicitly check this orientation; placing foci on $(0, \pm 3)$ when the major axis is horizontal costs the A1 even if $c$ is computed correctly. Diagnostic move: write "$a^{2} = $ larger denominator" before computing anything, this locks the major-axis orientation in.
Q7HARDHonors 🇨🇦 BC BC Provincial-style §4 Hyperbola, Foci & Asymptotes · HSG-GPE.A.3 (+) [9 marks]

Hyperbola $\tfrac{x^{2}}{9} - \tfrac{y^{2}}{16} = 1$. (a) $a$, $b$, transverse axis. (b) Vertices. (c) Foci via $c^{2} = a^{2} + b^{2}$. (d) Asymptotes. (e) Sign-difference vs ellipse.

Answer:  (a) $a = 3$, $b = 4$, horizontal transverse axis  ·  (b) $(\pm 3, 0)$  ·  (c) $c = 5$, foci $(\pm 5, 0)$  ·  (d) $y = \pm \tfrac{4}{3} x$  ·  (e) hyperbola foci lie outside the vertices ($c > a$), ellipse foci lie inside ($c < a$)

(a) Identify $a$, $b$, orientation A1·A1

The $x^{2}$ term is positive and the $y^{2}$ term is subtracted: standard form for a horizontal-transverse-axis hyperbola is $\dfrac{x^{2}}{a^{2}} - \dfrac{y^{2}}{b^{2}} = 1$, so $a^{2} = 9 \Rightarrow a = 3$ and $b^{2} = 16 \Rightarrow b = 4$. Transverse axis: horizontal (along the $x$-axis).

(b) Vertices A1

Endpoints of the transverse axis: $(\pm a, 0) = (\pm 3, 0)$.

(c) Foci via hyperbola focal relation M1·A1·A1

$c^{2} = a^{2} + b^{2} = 9 + 16 = 25 \Rightarrow c = 5$. Note the sign flip from the ellipse formula. Foci lie on the transverse axis: $F_{1, 2} = (\pm 5, 0)$.

(d) Asymptotes A1·A1

For a horizontal-transverse-axis hyperbola centred at the origin, the asymptotes are $y = \pm \dfrac{b}{a} x = \pm \dfrac{4}{3} x$.

(e) Why the focal-relation sign differs R1

An ellipse is defined by a constant sum of focal distances, $|PF_{1}| + |PF_{2}| = 2 a$, which forces $c < a$ (otherwise the triangle inequality fails) and gives $c^{2} = a^{2} - b^{2}$. A hyperbola is defined by a constant difference, $\bigl||PF_{1}| - |PF_{2}|\bigr| = 2 a$, which only requires $c > a$ (foci outside the vertices) and gives $c^{2} = a^{2} + b^{2}$. Sum-vs-difference of distances is the geometric reason for the sign flip.
Hyperbola: $c > a$ always; ellipse: $c < a$ always. The placement of the foci sorts the two conics instantly. For the ellipse, foci sit inside the closed curve, the "string and two thumbtacks" construction wraps around them. For the hyperbola, foci sit outside each branch, beyond the vertices, the asymptotes carry the branches away to infinity. If you ever compute $c > a$ on an ellipse problem or $c < a$ on a hyperbola problem, the algebra is wrong, audit immediately. BC PC-extension markers (and any first-year analytic-geometry course) reward students who lead with the conic's foci-vs-vertices ordering before crunching numbers.
Q8HARDHonors 🇨🇦 ON ON Provincial-style §5 Conic Discriminant $B^{2} - 4AC$ · HSG-GPE.A.3 (+) [9 marks]

Classify each general second-degree equation via $\Delta_{\text{conic}} = B^{2} - 4AC$. (a) $4x^{2} + 9y^{2} - \ldots$. (b) $x^{2} - y^{2} + \ldots$. (c) $y^{2} - 4x - \ldots$. (d) Circle test on (a).

Answer:  (a) $A = 4, B = 0, C = 9$, $\Delta = -144 < 0$, ellipse  ·  (b) $A = 1, B = 0, C = -1$, $\Delta = 4 > 0$, hyperbola  ·  (c) $A = 0, B = 0, C = 1$, $\Delta = 0$, parabola  ·  (d) not a circle since $A \ne C$

(a) Classify $4x^{2} + 9y^{2} - 16x + 18y - 11 = 0$ M1·A1·A1

Read coefficients: $A = 4$ (coefficient of $x^{2}$), $B = 0$ (no $xy$ term), $C = 9$ (coefficient of $y^{2}$). $$ \Delta_{\text{conic}} \;=\; B^{2} - 4 A C \;=\; 0 - 4(4)(9) \;=\; -144 \;<\; 0. $$ Since $\Delta_{\text{conic}} < 0$, the conic is an ellipse.

(b) Classify $x^{2} - y^{2} + 6x - 4y + 4 = 0$ M1·A1

$A = 1$, $B = 0$, $C = -1$. $\Delta_{\text{conic}} = 0 - 4(1)(-1) = 4 > 0 \Rightarrow$ hyperbola.

(c) Classify $y^{2} - 4x - 8y + 20 = 0$ M1·A1

No $x^{2}$ term, so $A = 0$; no $xy$ term, so $B = 0$; $C = 1$. $\Delta_{\text{conic}} = 0 - 4(0)(1) = 0 \Rightarrow$ parabola. (Cross-check: only one squared variable in the equation, $y^{2}$, with the other appearing linearly, the hallmark of a sideways-opening parabola.)

(d) Circle reduction test on (a) M1·A1

A circle is the special case of an ellipse with $A = C$ and $B = 0$. In (a), $A = 4$ but $C = 9$, so $A \ne C$ and the conic is not a circle, it is a genuine ellipse (the $x$- and $y$-radii differ). (Geometrically: completing the square gives $\tfrac{(x - 2)^{2}}{9} + \tfrac{(y + 1)^{2}}{4} = 1$, with $a^{2} = 9 \ne 4 = b^{2}$.)
$\Delta_{\text{conic}} = B^{2} - 4 A C$ is the conic analogue of the quadratic discriminant. Both are dimensionless tests that sort algebraic objects into qualitative classes by sign. For a single-variable quadratic, $\Delta = b^{2} - 4 a c$ sorts real/repeated/complex roots; for a two-variable conic, $\Delta_{\text{conic}} = B^{2} - 4 A C$ sorts ellipse/parabola/hyperbola. The structural parallel ($B^{2}$ on top, $4 A C$ on bottom of the test) is not a coincidence, both come from the same "completing the square" geometry. The $A = C, B = 0$ refinement is the circle's "perfect-square trinomial" analogue: a degenerate (more symmetric) case of the larger family. ON MHF4U does not require this classification but first-year university analytic geometry leans on it constantly.
Q9MEDIUM 🇺🇸 US AP-feeder FRQ §6 Translations of Conics · HSG-GPE.A.1 / A.3 [9 marks]

$x^{2} + y^{2} - 6x + 8y - 11 = 0$. (a) Complete the square. (b) Centre + radius. (c) Position of $(7, -4)$. (d) Transformation from unit circle.

Answer:  (a) $(x - 3)^{2} + (y + 4)^{2} = 36$  ·  (b) centre $(3, -4)$, $r = 6$  ·  (c) inside (distance $4 < r = 6$)  ·  (d) scale by $6$, then translate by $(3, -4)$

(a) Complete the square in $x$ and $y$ M1·A1·A1·A1

Group: $$ (x^{2} - 6x) + (y^{2} + 8y) = 11. $$ Half of $-6$ is $-3$; square is $9$. Half of $8$ is $4$; square is $16$. Add-and-subtract on the left, or equivalently add to both sides: $$ (x^{2} - 6x + 9) + (y^{2} + 8y + 16) \;=\; 11 + 9 + 16 \;=\; 36, $$ so $(x - 3)^{2} + (y + 4)^{2} = 36$.

(b) Centre and radius A1·A1

Centre $(h, k) = (3, -4)$ (read $h = 3$ from $(x - 3)$ and $k = -4$ from $(y + 4) = (y - (-4))$). Radius $r = \sqrt{36} = 6$.

(c) Position of $(7, -4)$ M1·A1

Substitute into the standard form: $(7 - 3)^{2} + (-4 - (-4))^{2} = 16 + 0 = 16$. Compare to $r^{2} = 36$: since $16 < 36$, the point lies inside the circle (its distance from the centre, $\sqrt{16} = 4$, is less than $r = 6$).

(d) Transformation from the unit circle A1

The unit circle $x^{2} + y^{2} = 1$ becomes $(x - 3)^{2} + (y + 4)^{2} = 36$ via two transformations: (i) scale (dilate) by factor $6$ about the origin, giving $x^{2} + y^{2} = 36$; (ii) translate by the vector $(3, -4)$, giving the final equation. (Equivalent order: translate first then scale about the translated centre, the result is the same.)
"Add-and-subtract the same number to both sides" $=$ complete the square geometry. The algebra of completing the square on $x^{2} + y^{2} + Dx + Ey + F = 0$ is identical to the algebra in Unit 2 (single-variable parabola CTS); the only new piece in Unit 12 is doing it in two variables in parallel. The resulting standard form $(x - h)^{2} + (y - k)^{2} = r^{2}$ then exposes the conic's geometry: $(h, k)$ is the centre (a translation) and $r$ is the scale factor relative to the unit circle. This unifies the translation language of Unit 5 (function transformations) with the analytic-geometry language of Unit 12, the AP graders explicitly look for that connection on HSG-GPE.A.1 questions.
PART III  ·  MODELING / APPLIED · SOLUTIONSUniversal · 28 marks

Section C · Modeling and Applications

Q10MEDIUM 🇺🇸 US AP-feeder FRQ §7 Parabolic Dish (Reflector) · HSG-GPE.A.2 [9 marks]

Dish: rim diameter $1.2$ m, depth $0.15$ m, $y = a x^{2}$ form with vertex at origin, axis along $+y$. (a) Rim edge coords. (b) Solve for $a$. (c) Focal length $p$ via $4p = 1/a$. (d) Receiver height + reflective property.

Answer:  (a) $(0.6, 0.15)$  ·  (b) $a = \tfrac{0.15}{0.36} = \tfrac{5}{12} \approx 0.4167$  ·  (c) $p = \tfrac{1}{4 a} = \tfrac{0.36}{0.6} = 0.60$ m  ·  (d) $60$ cm above vertex

(a) Rim edge coordinates A1

Rim diameter $1.2$ m means rim radius $0.6$ m horizontally. Depth $0.15$ m means the rim sits $0.15$ m above the vertex (the lowest point of the dish). So the rim edge directly above the rim's outer radius is at $(0.6, 0.15)$ in the chosen coordinates. (By symmetry the opposite edge is at $(-0.6, 0.15)$.)

(b) Solve for $a$ M1·A1

Substitute $(0.6, 0.15)$ into $y = a x^{2}$: $$ 0.15 \;=\; a (0.6)^{2} \;=\; 0.36\, a \;\Longrightarrow\; a \;=\; \frac{0.15}{0.36} \;=\; \frac{15}{36} \;=\; \frac{5}{12} \;\approx\; 0.4167. $$

(c) Focal length $p$ via $4p = 1/a$ M1·A1·A1

$$ p \;=\; \frac{1}{4 a} \;=\; \frac{1}{4 \cdot (5/12)} \;=\; \frac{12}{20} \;=\; \frac{3}{5} \;=\; 0.60 \;\text{m}. $$ Rounded to the nearest centimetre: $p \approx 0.60$ m $= 60$ cm.

(d) Receiver mount height + reflective property A1·A1·A1

The receiver sits at the focus, which is directly above the vertex on the axis of symmetry at height $p = 60$ cm $= 0.60$ m above the vertex of the dish. The reflective property of a parabola says that any ray travelling parallel to the axis of symmetry, after reflecting off the parabolic surface, passes through the focus. So parallel incoming radio waves (the standard assumption for a distant satellite signal) all converge at the focus, that is precisely why receivers on parabolic dishes are mounted there.
$4p = 1/a$ converts engineering geometry (depth + diameter) into parabolic optics (focal length) in one line. The engineering problem is given to you as rim diameter and depth; the optics problem requires the focal length. The bridge is $y = a x^{2}$ with one known point on the rim, then $4 p = 1/a$. Notice the answer here ($p = 60$ cm for a $120$-cm-diameter dish with $15$-cm depth) is realistic for a residential satellite dish, the "deeper, narrower" dishes have small $p$, and the "shallower, wider" dishes have large $p$. AP-feeder questions on HSG-GPE.A.2 increasingly reward students who name the reflective property explicitly, not just compute the focal length.
Q11MEDIUMHonors 🇨🇦 AB 🇺🇸 US AB Provincial-style §7 Kepler Orbit / Whisper Gallery · HSG-GPE.A.3 (+) [9 marks]

Whisper gallery: $24$ m long (major axis), $7$ m tall (semi-minor). (a) $a, b$. (b) $c$ via $c^{2} = a^{2} - b^{2}$. (c) Foci + separation. (d) Eccentricity + qualitative range.

Answer:  (a) $a = 12$, $b = 7$  ·  (b) $c = \sqrt{95} \approx 9.7$ m  ·  (c) foci $(\pm 9.7, 0)$, separation $\approx 19.5$ m  ·  (d) $e \approx 0.812$, strongly elongated

(a) Identify $a$ and $b$ A1

Major axis length $2a = 24 \Rightarrow a = 12$ m. Semi-minor axis $b = 7$ m (given).

(b) Focal distance $c$ M1·A1·A1

$$ c^{2} \;=\; a^{2} - b^{2} \;=\; 144 - 49 \;=\; 95 \;\Longrightarrow\; c \;=\; \sqrt{95} \;\approx\; 9.7468 \;\approx\; 9.7 \;\text{m}. $$

(c) Foci coordinates + separation A1·A1

With the ellipse centred at the origin and major axis along the $x$-axis, the foci sit on the major axis at $(\pm c, 0) \approx (\pm 9.7, 0)$. The distance between them is $2 c \approx 19.5$ m, the two visitors stand about $19.5$ m apart.

(d) Eccentricity and qualitative range M1·A1·A1

$$ e \;=\; \frac{c}{a} \;=\; \frac{\sqrt{95}}{12} \;\approx\; \frac{9.747}{12} \;\approx\; 0.812. $$ Eccentricity guide: $e \to 0$ would be "nearly circular" (foci almost coincide, no usable whisper effect because echoes return everywhere); $e \to 1$ is "strongly elongated" (foci far apart, sharp whisper effect because the elliptical reflection focuses precisely from one focus to the other). With $e \approx 0.81$, this gallery is strongly elongated and gives a clear whisper effect.
The whisper gallery is the reflective property of an ellipse, mirroring the dish's parabolic reflector. Where a parabola sends parallel rays to a single focus (Q10), an ellipse sends rays from one focus to the other focus. That is why two visitors at the foci of an elliptical room can whisper to each other: every sound ray leaving focus $F_{1}$ reflects off the ceiling and arrives at $F_{2}$. Kepler's first law (planets orbit on ellipses with the Sun at one focus) is the same geometry applied to gravity: the Sun is at one focus, the "empty" focus is just a geometric point. The eccentricity $e$ then tells you the orbital shape, Earth's $e \approx 0.017$ (nearly circular), Halley's comet $e \approx 0.97$ (extremely elongated). Honors-extension territory in AB Math 20-1 / 30-1 but core to AP Pre-Calculus and first-year astronomy.
Q12HARDHonors 🇨🇦 BC 🇺🇸 US BC Provincial-style §7 LORAN Hyperbolic Navigation · HSG-GPE.A.3 (+) [10 marks]

LORAN: stations $F_{1}(-200, 0)$, $F_{2}(200, 0)$ in km; ship's distance to $F_{1}$ exceeds distance to $F_{2}$ by $240$ km. (a) $c$ + $a$. (b) $b^{2}$. (c) Standard form. (d) Which branch. (e) Two-pair fix. (f) Asymptotes.

Answer:  (a) $c = 200$, $2a = 240 \Rightarrow a = 120$  ·  (b) $b^{2} = 25600$  ·  (c) $\tfrac{x^{2}}{14400} - \tfrac{y^{2}}{25600} = 1$  ·  (d) right branch ($x \ge 120$)  ·  (e) intersection of two hyperbolas  ·  (f) $y \approx \pm 1.333\, x$

(a) Focal distance $c$ and semi-transverse axis $a$ A1·A1

Stations sit at $(\pm 200, 0)$, so $c = 200$ km (centre-to-focus). The hyperbola is defined by the constant difference of focal distances: $\bigl||PF_{1}| - |PF_{2}|\bigr| = 2a$. Given that this difference is $240$ km: $2a = 240 \Rightarrow a = 120$ km. (Sanity check: $c > a$ as required for a hyperbola, $200 > 120$. $\checkmark$)

(b) $b^{2}$ via hyperbola focal relation M1·A1

$c^{2} = a^{2} + b^{2} \Rightarrow b^{2} = c^{2} - a^{2} = 200^{2} - 120^{2} = 40000 - 14400 = 25600$.

(c) Standard form of the hyperbola A1

With horizontal transverse axis and centre at the origin: $$ \frac{x^{2}}{14400} - \frac{y^{2}}{25600} \;=\; 1. $$

(d) Which branch M1·A1

$|PF_{1}| - |PF_{2}| = +240 > 0$, so the ship is farther from $F_{1}(-200, 0)$ than from $F_{2}(200, 0)$, i.e. closer to $F_{2}$. The right branch has $x \ge a = 120$ and sits on the same side of the centre as $F_{2}(+200, 0)$, so every point on the right branch is closer to $F_{2}$ than to $F_{1}$. Therefore the ship lies on the right branch ($x \ge 120$).

(e) Two-pair fix A1

A single hyperbola constrains the ship to one curve, infinitely many possible positions. A second independent pair of LORAN stations gives a second hyperbola, and the ship's exact position is the (unique nearby) intersection of the two hyperbolas. (In practice multiple intersections may arise globally; the ship's approximate position is used to pick the right one.)

(f) Asymptotes M1·A1

For $\dfrac{x^{2}}{a^{2}} - \dfrac{y^{2}}{b^{2}} = 1$, asymptotes are $y = \pm \dfrac{b}{a} x$: $$ \frac{b}{a} \;=\; \frac{\sqrt{25600}}{120} \;=\; \frac{160}{120} \;=\; \frac{4}{3} \;\approx\; 1.333. $$ Asymptotes: $y = \pm \dfrac{4}{3} x \approx \pm 1.333\, x$.
LORAN is the hyperbola's defining property turned into engineering. The hyperbola is defined as the locus of points with constant difference of distances to two foci. Time-of-arrival differences between two synchronized transmitters translate directly into distance differences (multiply by the speed of light), so a single pair of transmitters constrains a ship to one hyperbola; a second pair constrains it to another; the intersection fixes the ship's position. This is the navigation analogue of the ellipse's whisper gallery (Q11) and the parabola's reflector (Q10), each conic's defining property generates a real-world application. LORAN was deployed by the US Navy in WWII and stayed operational into the GPS era; the math is pure HSG-GPE.A.3 (+). The 3-4-5 triple sneaking out of $\sqrt{25600}/120 = 160/120 = 4/3$ is the same Pythagorean-triple shortcut that powered Q4's circle problem, both come from $c^{2} = a^{2} \pm b^{2}$ being a disguised Pythagorean theorem (right triangle inside the conic).