Hyperbola

Type 1

$$ \boxed{ \text{General formula: } {(x - h)^2 \over a^2} - {(y - k)^2 \over b^2} = 1 } $$

Features:

• Centre $(h, k)$
• Vertices $(h - a, k)$, $(h + a, k)$
• Two oblique asymptotes, $y = \pm{b \over a}(x - h) + k$
• $x$-values restricted to $x \le h - a$ or $x \ge h + a$

$$ {(x - 1)^2 \over 2^2} - {y^2 \over 1^2} = 1 $$

Features:

• Centre $(1, 0)$
• Vertices $(-1, 0)$, $(3, 0)$
• Two oblique asymptotes, $y = \pm {1 \over 2}(x - 1)$
• $x$-values restricted to $x \le -1$ or $x \ge 3$

Type 2

$$ \boxed{ \text{General formula: } {(y - k)^2 \over b^2} - {(x - h)^2 \over a^2} = 1 } $$

Features:

• Centre $(h, k)$
• Vertices $(h, k - b)$, $(h, k + b)$
• Two oblique asymptotes, $y = \pm{b \over a}(x - h) + k$
• $y$-values restricted to $y \le k - b$ or $y \ge k + b$

$$ {(y - 2)^2 \over 2^2} - {(x + 1)^2 \over 1^2} = 1 $$

Features:

• Centre $(-1, 2)$
• Vertices $(-1, 4)$ and $(-1, 0)$
• Two oblique asymptotes, $y = 2x + 4$ and $y = -2x$
• $y$-values restricted to $y \le 0$ or $y \ge 4$

Hyperbola in Conics app (TI-84 Plus CE)

$$ {x^2 \over 2^2} - (y - 1)^2 = 1 $$

To plot the hyperbola above in the Conics app:

1. Enter the Conics app ('apps' - 'Conics')
2. Select '3: Hyperbola'
3. Select the first hyperbola and enter the corresponding values:

'Top' & 'Bottom' Half, 'Left' & 'Right' Half

$$ {(y - 2)^2 \over 2^2} - (x + 1)^2 = 1 $$

To obtain the equation for the 'top' half and 'bottom' half, make $y$ the subject:

\begin{align} { (y - 2)^2 \over 2^2} - (x + 1)^2 & = 1 \\ { (y - 2)^2 \over 4} & = 1 + (x + 1)^2 \\ (y - 2)^2 & = 4 + 4(x + 1)^2 \\ y - 2 & = \pm \sqrt{4 + 4(x + 1)^2} \\ y & = \pm \sqrt{4 + 4(x + 1)^2} + 2 \end{align}

$$ {(y - 2)^2 \over 2^2} - (x + 1)^2 = 1 $$

On the other hand, to obtain the equation for the 'left' half and 'right' half, make $x$ the subject of the equation:

\begin{align} { (y - 2)^2 \over 2^2} - (x + 1)^2 & = 1 \\ { (y - 2)^2 \over 4} - (x + 1)^2 & = 1 \\ -(x + 1)^2 & = 1 - { (y - 2)^2 \over 4} \\ (x + 1)^2 & = { (y - 2)^2 \over 4} - 1 \\ x + 1 & = \pm \sqrt{ { (y - 2)^2 \over 4} - 1} \\ x & = \pm \sqrt{ { (y - 2)^2 \over 4} - 1} - 1 \\ x & = \pm \sqrt{ {1 \over 4}(y - 2)^2 - 1} - 1 \end{align}