Unit 9 · Conic Sections　圆锥曲线

By the end of this chapter you can:

1. Write the standard equation of an ellipse, hyperbola, or parabola and identify all key parameters from it
2. Locate foci, vertices, directrix, and asymptotes from the equation alone
3. Apply the focal distance identities — $|PF_1|+|PF_2|=2a$ (ellipse), $||PF_1|-|PF_2||=2a$ (hyperbola), focal-distance formula for a parabola — to compute lengths without coordinates
4. Classify any conic by its eccentricity and explain how $e$ controls the shape
5. Solve exam problems that mix two conics (shared foci, focal chords, eccentricity constraints)

Exam weight on past CSCA papers: ~12% (5–6 of 48 MCQs).

9.1　Ellipse　椭圆

Definition and standard equation

An ellipse is the locus of all points $P$ in the plane such that the sum of distances from $P$ to two fixed points $F_1$ and $F_2$ (the foci) is constant:

$$|PF_1| + |PF_2| = 2a \qquad (a > 0).$$

When the foci lie on the $x$-axis at $(\pm c,\ 0)$, the standard equation is:

$$\boxed{\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1} \qquad a > b > 0,\quad c^2 = a^2 - b^2.$$

When the foci lie on the $y$-axis at $(0,\ \pm c)$, swap $a^2$ and $b^2$ (i.e. $a^2$ sits under $y^2$). In either case $a$ is always the larger denominator.

Key parameters at a glance

🔑 The one identity to nail $c^2$: For an ellipse, $c^2 = a^2 - b^2$. For a hyperbola (Section 9.2), $c^2 = a^2 + b^2$. Many students mix these up — the ellipse subtracts because $c < a$.

⚠️ Exam trap: Always check which axis holds the foci. If the bigger denominator is under $y^2$, foci are on the $y$-axis.

Worked Example 9.1.A

The ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$ — find the foci, eccentricity, and the distance from point $P = (0, 3)$ to each focus.

Solution.

Read off: $a^2 = 25$, $b^2 = 9$, so $a = 5$, $b = 3$.

$$c = \sqrt{a^2 - b^2} = \sqrt{25 - 9} = \sqrt{16} = 4.$$

Foci are at $(\pm 4,\ 0)$.

$$e = \frac{c}{a} = \frac{4}{5} = 0.8.$$

The point $P = (0, 3)$ is a co-vertex. Use the focal-sum identity:

$$|PF_1| + |PF_2| = 2a = 10.$$

By symmetry, $|PF_1| = |PF_2|$, so each distance equals $5$.

Check: $\sqrt{(0-(-4))^2 + (3-0)^2} = \sqrt{16+9} = 5$. ✓

$$\boxed{|PF_1| = |PF_2| = 5,\quad e = 0.8}$$

9.2　Hyperbola　双曲线

Definition and standard equation

A hyperbola is the locus of all points $P$ such that the absolute difference of distances from $P$ to two fixed foci is constant:

$$\bigl||PF_1| - |PF_2|\bigr| = 2a \qquad (a > 0).$$

Standard equation with foci on the $x$-axis:

$$\boxed{\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1} \qquad a,\, b > 0,\quad c^2 = a^2 + b^2.$$

Foci are at $(\pm c,\ 0)$. The two branches open left and right.

With foci on the $y$-axis: $\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$.

Asymptotes　渐近线

The hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ approaches, but never touches, the lines:

$$y = \pm \frac{b}{a}\,x.$$

🔑 Trick to write asymptotes instantly: replace the $1$ on the right with $0$: $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 0 \;\Rightarrow\; y = \pm\dfrac{b}{a}x$.

Key parameters

⚠️ Common error: On a hyperbola, point $P$ must be on a specific branch. If $P$ is on the branch closer to $F_1$, then $|PF_1| < |PF_2|$ and $|PF_2| - |PF_1| = 2a$.

Worked Example 9.2.A

The hyperbola $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$: find the foci, asymptotes, and eccentricity.

Solution.

$a^2 = 16$, $b^2 = 9 \Rightarrow a = 4$, $b = 3$.

$$c = \sqrt{16 + 9} = \sqrt{25} = 5. \qquad \text{Foci: } (\pm 5,\ 0).$$

$$\text{Asymptotes: } y = \pm \frac{3}{4}\,x.$$

$$e = \frac{c}{a} = \frac{5}{4} = 1.25.$$

$$\boxed{F_{1,2} = (\pm 5,\ 0),\quad y = \pm\tfrac{3}{4}\,x,\quad e = \tfrac{5}{4}}$$

9.3　Parabola　抛物线

Definition

A parabola is the locus of all points $P$ equidistant from a fixed point (the focus $F$) and a fixed line (the directrix $l$):

$$|PF| = d(P,\, l).$$

Four standard orientations

🔑 Reading the parameter: For $y^2 = 4px$, the coefficient of $x$ is $4p$, so $p = \text{coefficient}/4$. For example, $y^2 = 12x$ gives $4p = 12$, $p = 3$, focus at $(3, 0)$, directrix $x = -3$.

⚠️ Vertex is at the origin for all standard forms above. If vertex is elsewhere, the equation shifts — but CSCA questions almost always use vertex at the origin.

Worked Example 9.3.A

Find the focus and directrix of $y^2 = -8x$, then find the distance from $P = (-2, 4)$ to the focus.

Solution.

This is $y^2 = -4px$ with $4p = 8 \Rightarrow p = 2$. Opens left.

Focus: $(-2,\ 0)$. Directrix: $x = 2$.

Distance to focus: $|PF| = \sqrt{(-2-(-2))^2 + (4-0)^2} = \sqrt{0+16} = 4$.

Verify with directrix definition: $d(P,\; x=2) = |-2 - 2| = 4$. ✓

$$\boxed{\text{Focus } (-2,\,0),\quad \text{directrix } x = 2,\quad |PF| = 4}$$

9.4　Focal Properties　焦点性质

This section gathers the most tested calculation shortcuts.

Ellipse focal-distance formulas

For a point $P = (x_0, y_0)$ on the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ with foci $F_1 = (-c, 0)$ (left) and $F_2 = (c, 0)$ (right):

$$|PF_1| = a + ex_0 \qquad |PF_2| = a - ex_0 \qquad \text{where } e = \frac{c}{a}.$$

Since $-a \le x_0 \le a$, we have $0 \le |PF_1| \le 2a$ and $0 \le |PF_2| \le 2a$. The point is closest to the right focus when $x_0 > 0$.

Derivation sketch: From $|PF_1|+|PF_2|=2a$ and squaring the individual distance expressions, one obtains $|PF_1|-|PF_2| = 2ex_0$. Adding and subtracting gives the formulas above.

Parabola focal-distance formula

For $P = (x_0, y_0)$ on $y^2 = 4px$ ($p>0$, focus at $(p,0)$, directrix $x = -p$):

$$|PF| = x_0 + p$$

because the directrix is $x = -p$, and by definition $|PF| = d(P,\,l) = x_0 - (-p) = x_0 + p$.

🔑 Fastest CSCA trick: Instead of the distance formula (which needs a square root), just read off $x_0$ and add $p$.

Focal chord — latus rectum　通径

A focal chord is a chord of the conic passing through a focus. The special case perpendicular to the axis of symmetry is the latus rectum (通径):

• Parabola $y^2 = 4px$: latus rectum passes through $(p, 0)$; endpoints at $(p, \pm 2p)$; length $= 4p$.
• Ellipse: latus rectum half-length $= b^2/a$; full length $= 2b^2/a$.
• Hyperbola: latus rectum full length $= 2b^2/a$.

Worked Example 9.4.A

Point $P$ is on the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$. The right focus is $F_2$. If $|PF_2| = 3$, find $|PF_1|$.

Solution.

$a^2 = 25$, $b^2 = 16 \Rightarrow a = 5$.

$$|PF_1| + |PF_2| = 2a = 10 \implies |PF_1| = 10 - 3 = 7.$$

$$\boxed{|PF_1| = 7}$$

Worked Example 9.4.B

Point $P$ is on the parabola $y^2 = 12x$ and $|PF| = 7$ (where $F$ is the focus). Find the $x$-coordinate of $P$.

Solution.

$4p = 12 \Rightarrow p = 3$. So $|PF| = x_0 + p = x_0 + 3 = 7 \Rightarrow x_0 = 4$.

$$\boxed{x_0 = 4}$$

9.5　Eccentricity　离心率

The master classification

Eccentricity $e = c/a$ classifies the conic by shape:

$$e = 0 \;\Longrightarrow\; \text{circle} \qquad 0 < e < 1 \;\Longrightarrow\; \text{ellipse} \qquad e = 1 \;\Longrightarrow\; \text{parabola} \qquad e > 1 \;\Longrightarrow\; \text{hyperbola}$$

How eccentricity controls shape:

• Ellipse: as $e \to 0^+$, $c \to 0$ and the ellipse approaches a circle (both axes equal). As $e \to 1^-$, the ellipse becomes very elongated — almost a line segment.
• Hyperbola: as $e \to 1^+$, branches are nearly parabolic and open very narrowly. As $e \to \infty$, the asymptotes' angle approaches $90°$ — the branches open widely.
• Parabola: $e = 1$ is exact; it is the transition case between the two families.

⚠️ Exam favourite: "Given that two conics share the same foci, find the eccentricity of each." Remember: $c$ is the same for both (shared foci), but $a$ differs.

Worked Example 9.5.A

A hyperbola and an ellipse share foci at $(\pm\sqrt{5},\ 0)$. The ellipse has $a = 3$; the hyperbola has $a' = 2$. Find both eccentricities and the equations of the two conics.

Solution.

Shared $c = \sqrt{5}$.

Ellipse: $a = 3$, $b^2 = a^2 - c^2 = 9 - 5 = 4$.

$$\frac{x^2}{9} + \frac{y^2}{4} = 1, \qquad e_{\text{ell}} = \frac{\sqrt{5}}{3}.$$

Hyperbola: $a' = 2$, $b'^2 = c^2 - a'^2 = 5 - 4 = 1$.

$$\frac{x^2}{4} - y^2 = 1, \qquad e_{\text{hyp}} = \frac{\sqrt{5}}{2}.$$

$$\boxed{e_{\text{ellipse}} = \frac{\sqrt{5}}{3},\quad e_{\text{hyperbola}} = \frac{\sqrt{5}}{2}}$$

Summary comparison table — all three conics　三种圆锥曲线对比

Try it!　自测练习

Q1. Write the standard equation of the ellipse with foci at $(\pm 3,\ 0)$ and major-axis length $10$.

Q2. The hyperbola $\dfrac{x^2}{4} - \dfrac{y^2}{5} = 1$: find its foci, eccentricity, and asymptotes.

Q3. The parabola $x^2 = -12y$: state the focus and directrix.

Q4. Point $P$ is on the parabola $y^2 = 4x$ and $|PF| = 5$ (where $F$ is the focus). Find the $x$-coordinate of $P$.

Q5. An ellipse has equation $\dfrac{x^2}{9} + \dfrac{y^2}{5} = 1$. A point $P$ on the ellipse satisfies $|PF_1| = 1$. Find $|PF_2|$.

📌 Chapter summary

Topic	Key equation / fact
Ellipse	$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$; $c^2=a^2-b^2$; $\|PF_1\|+\|PF_2\|=2a$; $e\in(0,1)$
Hyperbola	$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$; $c^2=a^2+b^2$; $\bigl\|\|PF_1\|-\|PF_2\|\bigr\|=2a$; $e>1$; asymptotes $y=\pm\frac{b}{a}x$
Parabola	$y^2=4px$; focus $(p,0)$; directrix $x=-p$; $\|PF\|=x_0+p$; $e=1$
Focal distances	Ellipse: $\|PF\|=a\pm ex_0$. Parabola: add $p$ to the $x$-coordinate of $P$.
Eccentricity	$e<1$: ellipse; $e=1$: parabola; $e>1$: hyperbola. Closer to 0 = rounder ellipse.
Exam tip	Always identify the conic type first, then extract $a, b, c$ — and never mix up $c^2 = a^2-b^2$ (ellipse) vs $c^2 = a^2+b^2$ (hyperbola).

What's next → Unit 10 covers Complex Numbers — a shorter unit (~4%) but a reliable quick win. The modulus $|z| = \sqrt{a^2+b^2}$ echoes the focal-distance geometry you just practised.