The Logic of the Hyperbola#
A hyperbola is what happens when you divide by $x$. The function $y = \frac{1}{x}$ can never equal zero (you can’t divide something and get nothing), and $x$ can never be zero (you can’t divide by nothing). This creates a graph with two separate branches and two invisible boundary lines called asymptotes.
1. The General Form#
$$ y = \frac{a}{x - p} + q $$This is the form you must know for Grade 12. Every hyperbola question can be answered using this equation.
2. The Parameters: What Each One Does#
The “$a$” Value — Branch Size and Orientation#
The $a$ value controls two things: how “fat” the branches are and which quadrants they sit in.
| Value of $a$ | Effect |
|---|---|
| $a > 0$ | Branches are in Quadrants I and III (top-right and bottom-left) |
| $a < 0$ | Branches are in Quadrants II and IV (top-left and bottom-right) |
| $ | a |
| $ | a |
Think of $a$ as a “flip switch” and a “zoom”. Positive $a$ = standard orientation. Negative $a$ = flipped diagonally.
The “$p$” Value — Horizontal Shift (Vertical Asymptote)#
| Value of $p$ | Effect |
|---|---|
| $p > 0$ | Graph shifts right by $p$ units |
| $p < 0$ | Graph shifts left by $ |
| $p = 0$ | Graph is centred on the y-axis |
The vertical asymptote is the line $x = p$.
The graph can never touch or cross this line. It represents the value of $x$ that would make the denominator zero (which is impossible).
The “opposite sign” trap (same as the parabola): In $y = \frac{3}{x + 2} + 1$, we have $p = -2$ (shift left 2), because $x + 2 = x - (-2)$.
The “$q$” Value — Vertical Shift (Horizontal Asymptote)#
| Value of $q$ | Effect |
|---|---|
| $q > 0$ | Graph shifts up by $q$ units |
| $q < 0$ | Graph shifts down by $ |
| $q = 0$ | Horizontal asymptote is the x-axis |
The horizontal asymptote is the line $y = q$.
As $x$ gets very large (positive or negative), $\frac{a}{x - p}$ gets closer and closer to zero, so $y$ approaches $q$ but never reaches it.
The Centre of the Hyperbola#
The point where the two asymptotes cross is the centre of the hyperbola: $(p; q)$.
This is not a point on the graph — it’s the “invisible centre” that the two branches curve around.
3. Key Properties#
| Property | Value |
|---|---|
| Vertical Asymptote | $x = p$ |
| Horizontal Asymptote | $y = q$ |
| Centre | $(p; q)$ |
| Domain | $x \in \mathbb{R}, x \ne p$ |
| Range | $y \in \mathbb{R}, y \ne q$ |
| y-intercept | Set $x = 0$: $y = \frac{a}{0 - p} + q = -\frac{a}{p} + q$ |
| x-intercept | Set $y = 0$: $0 = \frac{a}{x - p} + q$, so $x = p - \frac{a}{q}$ |
| Axes of Symmetry | $y = x - p + q$ and $y = -x + p + q$ (diagonal lines through the centre) |
4. Sketching a Hyperbola — Step by Step#
Example: Sketch $y = \frac{2}{x - 1} + 3$
- Identify the parameters: $a = 2$, $p = 1$, $q = 3$.
- Draw the asymptotes:
- Vertical: $x = 1$ (dashed line)
- Horizontal: $y = 3$ (dashed line)
- Mark the centre: $(1; 3)$
- Determine the orientation: $a = 2 > 0$, so branches are in Quadrants I and III relative to the centre.
- Find the intercepts:
- y-intercept: $y = \frac{2}{0 - 1} + 3 = -2 + 3 = 1$. Point: $(0; 1)$
- x-intercept: $0 = \frac{2}{x - 1} + 3 \Rightarrow \frac{2}{x-1} = -3 \Rightarrow x - 1 = -\frac{2}{3} \Rightarrow x = \frac{1}{3}$. Point: $(\frac{1}{3}; 0)$
- Plot the points and sketch the two smooth branches curving toward the asymptotes.
5. Finding the Equation from a Graph#
Given the asymptotes and one point#
Example: Asymptotes at $x = -2$ and $y = 4$, graph passes through $(0; 5)$.
- From the asymptotes: $p = -2$, $q = 4$.
- Write: $y = \frac{a}{x - (-2)} + 4 = \frac{a}{x + 2} + 4$
- Substitute $(0; 5)$: $$ 5 = \frac{a}{0 + 2} + 4 $$ $$ 1 = \frac{a}{2} $$ $$ a = 2 $$
- Final equation: $y = \frac{2}{x + 2} + 4$
Given the axes of symmetry#
If you are given $y = x + 1$ and $y = -x + 5$ as axes of symmetry:
- The centre is where they intersect. Solve simultaneously: $$ x + 1 = -x + 5 $$ $$ 2x = 4 $$ $$ x = 2, \quad y = 3 $$
- Centre: $(2; 3)$, so $p = 2$, $q = 3$.
- Use a point on the graph to find $a$.
6. The Inverse of a Hyperbola#
The inverse of a hyperbola is another hyperbola (reflected across $y = x$).
For the basic hyperbola $y = \frac{a}{x}$:
- Swap: $x = \frac{a}{y}$
- Solve: $y = \frac{a}{x}$
The basic hyperbola is its own inverse! This makes sense because $y = \frac{a}{x}$ is symmetric about $y = x$.
For the shifted form $y = \frac{a}{x - p} + q$, the inverse has its asymptotes swapped: the vertical asymptote becomes horizontal and vice versa.
7. Domain and Range Questions#
Exam questions often ask: “For which values of $x$ is $f(x) > 0$?” or “For which values of $x$ is $f(x) \ge q$?”
Strategy:
- Sketch the graph (even roughly).
- Find the x-intercept.
- Read the answer from the graph, remembering to exclude the asymptote.
Example: For $y = \frac{2}{x - 1} + 3$, for which values of $x$ is $f(x) > 3$?
Since $y = 3$ is the horizontal asymptote and $a > 0$:
- The branch above the asymptote ($f(x) > 3$) is in the region $x > 1$.
- Answer: $x > 1$
Worked Example: Full Exam-Style Question#
Given $g(x) = \frac{-4}{x + 3} - 1$
(a) Write down the equations of the asymptotes.
- Vertical asymptote: $x = -3$
- Horizontal asymptote: $y = -1$
(b) Determine the intercepts.
y-intercept ($x = 0$):
$$ g(0) = \frac{-4}{0 + 3} - 1 = -\frac{4}{3} - 1 = -\frac{7}{3} \approx -2.33 $$x-intercept ($y = 0$):
$$ 0 = \frac{-4}{x + 3} - 1 $$$$ \frac{-4}{x + 3} = 1 $$$$ -4 = x + 3 $$$$ x = -7 $$(c) Write down the domain and range.
- Domain: $x \in \mathbb{R}, x \ne -3$
- Range: $y \in \mathbb{R}, y \ne -1$
(d) For which values of $x$ is $g(x) < -1$?
$y = -1$ is the horizontal asymptote. Since $a = -4 < 0$, the branches are in Quadrants II and IV relative to the centre $(-3; -1)$.
The branch below the asymptote ($g(x) < -1$) is where $x > -3$.
Answer: $x > -3$
(e) Determine the equations of the axes of symmetry.
- $y = x - (-3) + (-1) = x + 3 - 1 = x + 2$
- $y = -(x - (-3)) + (-1) = -x - 3 - 1 = -x - 4$
🚨 Common Mistakes#
- Asymptotes are NOT intercepts: The asymptotes are invisible boundary lines. The graph never touches them. Don’t confuse them with where the graph crosses the axes.
- Forgetting $x \ne p$ in the domain: Many students write “Domain: $x \in \mathbb{R}$” and forget to exclude the vertical asymptote. This costs marks every time.
- Sign of $p$: In $y = \frac{3}{x + 5}$, the vertical asymptote is $x = -5$ (not $x = 5$).
- Quadrant confusion: When the centre shifts, the “quadrants” shift with it. A hyperbola with $a > 0$ always has branches in Quadrants I and III relative to its centre, not relative to the origin.
💡 Pro Tip: The Asymptote Equations#
If you forget everything else, remember this: the asymptotes are $x = p$ and $y = q$. If you can read the asymptotes from a graph, you immediately know two of the three parameters. Then use any point on the graph to find $a$.
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