椭圆tuǒyuán

Core Concept

An ellipse (椭圆) is the set of points whose distances to two fixed points (the foci, 焦点) have a constant sum:

$|PF_1| + |PF_2| = 2a \quad (2a > |F_1F_2|)$

The condition $2a > |F_1F_2|$ matters: if the sum equals the distance between the foci you get a line segment, and if it is smaller you get nothing at all. Exam questions test this boundary directly.

A circle is the special case where the two foci coincide.

Standard Equations

With centre at the origin and $a > b > 0$:

Equation	Foci on	Foci	Vertices
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$	$x$-axis	$(\pm c, 0)$	$(\pm a, 0)$
$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$	$y$-axis	$(0, \pm c)$	$(0, \pm a)$

How to tell them apart: the foci lie along whichever axis has the larger denominator. This is the most reliable rule on the topic — it does not depend on remembering which letter goes where.

The Fundamental Relationship

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

• $a$ = semi-major axis (长半轴)
• $b$ = semi-minor axis (短半轴)
• $c$ = focal distance from centre (半焦距)

Contrast with the hyperbola, where the relationship is $c^2 = a^2 + b^2$. Mixing these two up is the single most common conic-section error. A memory hook: for an ellipse, $a$ is the biggest (it is the largest denominator), so $a$ sits alone on the left.

Eccentricity

$e = \frac{c}{a}, \qquad 0 < e < 1$

As $e \to 0$ the ellipse becomes circular; as $e \to 1$ it becomes long and flat.

Key Properties

Property 1: The Focal Triangle

For a point $P$ on the ellipse, the triangle $PF_1F_2$ has a fixed perimeter:

$|PF_1| + |PF_2| + |F_1F_2| = 2a + 2c$

Questions about the perimeter of this triangle are effectively one-liners.

Property 2: Range of the Focal Radius

$a - c \leq |PF_1| \leq a + c$

The minimum occurs at the nearer vertex, the maximum at the farther one.

Property 3: Latus Rectum

The chord through a focus perpendicular to the major axis has length $\frac{2b^2}{a}$.

Real-World Application

Planetary orbits are ellipses with the sun at one focus (Kepler's first law). Earth's orbit has $e \approx 0.017$ — so close to zero that it is nearly circular, which is why the seasons are caused by axial tilt rather than by distance from the sun.

CSCA Practice Problems

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

Example 1: Basic (Difficulty ★★☆☆☆)

For the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$, the coordinates of the foci are:

Options:

• A. $(\pm 3, 0)$
• B. $(0, \pm 3)$
• C. $(\pm 41, 0)$
• D. $(\pm 9, 0)$

Detailed Solution:

1. The larger denominator is $25$, sitting under $x^2$ — so the foci are on the $x$-axis. Option B is eliminated.

2. Read off $a^2 = 25$ and $b^2 = 16$.

3. Apply $a^2 = b^2 + c^2$: $c^2 = a^2 - b^2 = 25 - 16 = 9 \implies c = 3$

4. Foci: $(\pm 3, 0)$.

Option D is the trap for stopping at $c^2 = 9$; option C is the trap for adding instead of subtracting (the hyperbola relationship).

Answer: A

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

An ellipse has foci $(\pm 4, 0)$ and passes through the point $(0, 3)$. Find its standard equation.

Options:

• A. $\frac{x^2}{25} + \frac{y^2}{9} = 1$
• B. $\frac{x^2}{9} + \frac{y^2}{25} = 1$
• C. $\frac{x^2}{16} + \frac{y^2}{9} = 1$
• D. $\frac{x^2}{25} + \frac{y^2}{16} = 1$

Detailed Solution:

1. The foci are on the $x$-axis, so the form is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $c = 4$.

2. The point $(0, 3)$ is on the ellipse. Substituting $x = 0$: $\frac{9}{b^2} = 1 \implies b^2 = 9$

   (Geometrically, $(0,3)$ is the endpoint of the minor axis, so $b = 3$ directly.)

3. Apply $a^2 = b^2 + c^2$: $a^2 = 9 + 16 = 25$

4. Equation: $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

5. Verify: $a = 5 > b = 3$ ✓, and the larger denominator is under $x^2$, matching foci on the $x$-axis ✓.

Answer: A

Common Mistakes

❌ Mistake 1: Using $c^2 = a^2 + b^2$

That is the hyperbola relationship. For an ellipse it is $a^2 = b^2 + c^2$, equivalently $c^2 = a^2 - b^2$. Since $a$ is the largest, $c$ must come from a subtraction.

❌ Mistake 2: Assuming $a^2$ is always under $x^2$

It is under whichever variable's axis carries the foci. In $\frac{x^2}{9} + \frac{y^2}{25} = 1$ the foci are on the $y$-axis, with $a^2 = 25$.

❌ Mistake 3: Reporting $c^2$ as the focal coordinate

If $c^2 = 9$, the foci are at $\pm 3$, not $\pm 9$.

❌ Mistake 4: Forgetting the $2a > 2c$ condition

When a question includes a parameter, check that the resulting $a$ genuinely exceeds $c$ — otherwise the "ellipse" does not exist.

Study Tips

1. ✅ Learn the "larger denominator marks the focal axis" rule — it removes all guesswork
2. ✅ Write $a^2 = b^2 + c^2$ and $c^2 = a^2 + b^2$ side by side and label which conic each belongs to
3. ✅ Extract $a$, $b$, $c$ first, before attempting anything else
4. ✅ Use the defining sum $|PF_1| + |PF_2| = 2a$ whenever both foci appear in a question
5. ✅ Check $a > b$ in your final answer as a self-test

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💡 Exam Tip: 椭圆定义及辨析 and 根据 $a$、$b$、$c$ 求椭圆标准方程 are both recurring CSCA knowledge points. Nearly every ellipse question reduces to correctly extracting $a$, $b$, $c$ and applying $a^2 = b^2 + c^2$.