离心率líxīnlǜ

Core Concept

The eccentricity (离心率) $e$ measures how far a conic section departs from being circular. It is the one number that classifies all conics on a single scale, and CSCA questions use it constantly as a bridge between $a$, $b$, and $c$.

Definition

$e = \frac{c}{a}$

where $c$ is the distance from the centre to a focus and $a$ is the semi-major axis (ellipse) or the real semi-axis (hyperbola).

The Classification Scale

Value	Conic	Chinese
$e = 0$	circle	圆
$0 < e < 1$	ellipse	椭圆
$e = 1$	parabola	抛物线
$e > 1$	hyperbola	双曲线

This single table is worth memorising outright — it converts many multiple-choice questions into a one-second lookup.

The Unified Definition

There is a second, deeper definition that explains why one number covers all four cases. For a fixed point $F$ (focus) and fixed line $l$ (directrix), the set of points $P$ satisfying

$\frac{|PF|}{d(P, l)} = e$

is a conic with eccentricity $e$. The parabola's defining property — equal distances to focus and directrix — is exactly the case $e = 1$.

Computing Eccentricity

For an Ellipse

Since $a^2 = b^2 + c^2$:

$e = \frac{c}{a} = \frac{\sqrt{a^2 - b^2}}{a} = \sqrt{1 - \frac{b^2}{a^2}}$

The last form is the most useful in exams: it lets you compute $e$ directly from the two denominators of the standard equation, with no need to find $c$ first.

For a Hyperbola

Since $c^2 = a^2 + b^2$:

$e = \frac{c}{a} = \frac{\sqrt{a^2 + b^2}}{a} = \sqrt{1 + \frac{b^2}{a^2}}$

Note the sign difference: minus for the ellipse, plus for the hyperbola. This mirrors the two fundamental relationships and is the usual place students go wrong.

Link to Asymptotes

For a hyperbola, the asymptote slope is $\frac{b}{a}$, so eccentricity and asymptote slope are two views of the same ratio:

$e^2 = 1 + \left(\frac{b}{a}\right)^2$

A question giving asymptotes therefore gives eccentricity for free, and vice versa.

CSCA Practice Problems

💡 Note: These problems follow the CSCA exam syllabus and Chinese standardized test formats.

Example 1: Basic (Difficulty ★★★☆☆)

The eccentricity of the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$ is:

Options:

• A. $\frac{3}{5}$
• B. $\frac{4}{5}$
• C. $\frac{5}{3}$
• D. $\frac{9}{25}$

Detailed Solution:

1. The larger denominator is under $x^2$, so $a^2 = 25$ and $b^2 = 16$, giving $a = 5$.

2. Compute $c$ from the ellipse relationship: $c^2 = a^2 - b^2 = 25 - 16 = 9 \implies c = 3$

3. Therefore: $e = \frac{c}{a} = \frac{3}{5}$

Faster route: $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$ — one step, no $c$ needed.

Option C is $\frac{a}{c}$ (inverted, and impossible since ellipse eccentricity must be below $1$); option D is $e^2$.

Answer: A

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Example 2: Intermediate (Difficulty ★★★★☆)

A hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ has asymptotes $y = \pm\frac{3}{4}x$. Find its eccentricity.

Options:

• A. $\frac{5}{4}$
• B. $\frac{5}{3}$
• C. $\frac{4}{3}$
• D. $\frac{\sqrt{7}}{4}$

Detailed Solution:

1. The asymptote slope for this orientation is $\frac{b}{a}$, so: $\frac{b}{a} = \frac{3}{4}$

2. Apply the hyperbola eccentricity formula: $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \left(\frac{3}{4}\right)^2} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$

3. Sanity check: $e = 1.25 > 1$ ✓ — consistent with a hyperbola.

Option D comes from using the ellipse formula ($\sqrt{1 - \frac{9}{16}} = \frac{\sqrt{7}}{4}$), which would also fail the $e > 1$ check.

Answer: A

Common Mistakes

❌ Mistake 1: Inverting the ratio

$e = \frac{c}{a}$, not $\frac{a}{c}$. A quick check: an ellipse must give $e < 1$ and a hyperbola $e > 1$. If your answer violates that, you inverted.

❌ Mistake 2: Using the wrong sign inside the square root

Ellipse: $\sqrt{1 - \frac{b^2}{a^2}}$. Hyperbola: $\sqrt{1 + \frac{b^2}{a^2}}$. Always confirm which conic you are on before choosing.

❌ Mistake 3: Reporting $e^2$

Squaring appears mid-derivation; remember to take the final square root.

❌ Mistake 4: Forgetting the range constraints

If a question asks "for what values of $m$ does this represent an ellipse with $e < \frac{1}{2}$", the answer must also satisfy the basic existence conditions ($a > b > 0$). Range questions are where eccentricity gets genuinely hard.

Study Tips

1. ✅ Memorise the classification table ($e = 0$ / $0<e<1$ / $e=1$ / $e>1$)
2. ✅ Use $e = \sqrt{1 \mp \frac{b^2}{a^2}}$ to skip computing $c$
3. ✅ Sanity-check your answer against the conic type — this catches inversions instantly
4. ✅ Connect $e$ to asymptote slope for hyperbolas: $e^2 = 1 + (b/a)^2$
5. ✅ Practice range problems, where $e$ is bounded and you solve for a parameter

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💡 Exam Tip: Eccentricity is the most common way CSCA links the three conics in one question. Because it packages $a$, $b$, and $c$ into a single number, questions often give $e$ and ask for the equation — or give the equation and ask for $e$.