双曲线shuāngqūxiàn

Core Concept

A hyperbola (双曲线) is the set of points whose distances to two fixed points (the foci) have a constant difference in absolute value:

$\left| |PF_1| - |PF_2| \right| = 2a \quad (0 < 2a < |F_1F_2|)$

Note the contrast with the ellipse, which uses a constant sum. The absolute-value bars matter: dropping them gives only one of the two branches.

The condition $2a < |F_1F_2|$ is the existence requirement — the reverse of the ellipse's condition.

Standard Equations

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

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

How to tell them apart: unlike the ellipse, size does not decide. The foci lie along the axis of the variable carrying the positive term. In $\frac{y^2}{9} - \frac{x^2}{16} = 1$ the positive term is $y^2$, so the foci are on the $y$-axis even though $16 > 9$.

The Fundamental Relationship

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

Here $c$ is the largest quantity, so $c$ sits alone. This is the exact opposite of the ellipse's $a^2 = b^2 + c^2$, and confusing the two is the most common conic error in the whole syllabus.

	Ellipse	Hyperbola
Defining condition	sum $= 2a$	|difference| $= 2a$
Relationship	$a^2 = b^2 + c^2$	$c^2 = a^2 + b^2$
Largest letter	$a$	$c$
Eccentricity	$0 < e < 1$	$e > 1$
Asymptotes	none	yes

Asymptotes

The defining feature of the hyperbola. For $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:

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

For $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$:

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

A reliable derivation: replace the $1$ on the right with $0$ and solve. From $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 0$ we get $\frac{y^2}{b^2} = \frac{x^2}{a^2}$, hence $y = \pm\frac{b}{a}x$. This trick works for both orientations, so you never need to memorise which fraction goes on top.

Eccentricity

$e = \frac{c}{a} > 1$

Larger $e$ means the branches open more widely.

CSCA Practice Problems

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

Example 1: Basic (Difficulty ★★★☆☆)

The foci of the hyperbola $\frac{x^2}{9} - \frac{y^2}{16} = 1$ are:

Options:

• A. $(\pm 5, 0)$
• B. $(0, \pm 5)$
• C. $(\pm \sqrt{7}, 0)$
• D. $(\pm 7, 0)$

Detailed Solution:

1. The positive term is $x^2$, so the foci lie on the $x$-axis. Option B is eliminated.

2. Read off $a^2 = 9$, $b^2 = 16$.

3. Apply $c^2 = a^2 + b^2$ (addition, not subtraction): $c^2 = 9 + 16 = 25 \implies c = 5$

4. Foci: $(\pm 5, 0)$.

Option C is the trap for students who subtract, as they would for an ellipse: $16 - 9 = 7$.

Answer: A

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

A hyperbola has asymptotes $y = \pm 2x$ and passes through $(2, 2\sqrt{3})$. Find its standard equation.

Options:

• A. $x^2 - \frac{y^2}{4} = 1$
• B. $\frac{x^2}{4} - y^2 = 1$
• C. $y^2 - \frac{x^2}{4} = 1$
• D. $\frac{x^2}{3} - \frac{y^2}{12} = 1$

Detailed Solution:

1. The point $(2, 2\sqrt{3})$ has $|x| = 2$, and a hyperbola's branches extend along the focal axis — so try foci on the $x$-axis: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

2. The asymptotes of this form are $y = \pm\frac{b}{a}x$. Matching $y = \pm 2x$: $\frac{b}{a} = 2 \implies b = 2a \implies b^2 = 4a^2$

3. Substitute the point $(2, 2\sqrt{3})$, using $b^2 = 4a^2$: $\frac{2^2}{a^2} - \frac{(2\sqrt{3})^2}{4a^2} = 1$ $\frac{4}{a^2} - \frac{12}{4a^2} = 1$

4. Simplify the second fraction, then combine over the common denominator $a^2$: $\frac{4}{a^2} - \frac{3}{a^2} = \frac{1}{a^2} = 1 \implies a^2 = 1$

5. Therefore $b^2 = 4a^2 = 4$, and the equation is: $x^2 - \frac{y^2}{4} = 1$

6. Verify both given conditions:

   • Asymptotes: $\frac{b}{a} = \frac{2}{1} = 2$ ✓
   • Point: $2^2 - \frac{12}{4} = 4 - 3 = 1$ ✓

Answer: A

Method note: Option D also has asymptote slope $\sqrt{12/3} = 2$, so it passes the first check — but substituting the point gives $\frac{4}{3} - 1 = \frac{1}{3} \neq 1$. Asymptotes alone never determine a hyperbola uniquely; they fix only the ratio $\frac{b}{a}$, so a second condition is always required.

Common Mistakes

❌ Mistake 1: Using the ellipse relationship

For a hyperbola $c^2 = a^2 + b^2$ — always addition. If you find yourself subtracting, you have slipped into ellipse mode.

❌ Mistake 2: Deciding the orientation by size

For an ellipse the larger denominator wins; for a hyperbola the positive term wins. These are different rules, and applying the ellipse rule here is a guaranteed error.

❌ Mistake 3: Inverting the asymptote slope

For $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ the slope is $\frac{b}{a}$, not $\frac{a}{b}$. Use the set-the-equation-to-zero trick and you cannot get this wrong.

❌ Mistake 4: Dropping the absolute value in the definition

$|PF_1| - |PF_2| = 2a$ describes a single branch; $\left||PF_1| - |PF_2|\right| = 2a$ describes both.

Study Tips

1. ✅ Build the ellipse-versus-hyperbola comparison table and review it before every practice session
2. ✅ Derive asymptotes by setting the right side to $0$ instead of memorising
3. ✅ Identify the positive term first to fix the orientation
4. ✅ Remember $c$ is largest for hyperbolas, $a$ is largest for ellipses
5. ✅ Substitute back into every given condition before choosing an option

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💡 Exam Tip: 求双曲线的焦点坐标 and 已知方程求双曲线的渐近线 are both recurring CSCA knowledge points. Asymptote questions are the more common of the two, and the set-to-zero derivation makes them nearly automatic.