Explore: Graphing Hyperbolas Solutions

Graphing Hyperbolas Solutions

To graph a hyperbola, follow these guidelines: 
- Write the equation in standard form (if needed). 
- Sketch the opening of the branches (recommended, but optional). 
- Determine/Mark the center . 
- Mark the vertices and co-vertices as you would for an ellipse. 
- Sketch a dashed rectangle to help graph the asymptotes (recommended, but optional). 
- Graph the asymptotes from the center to the vertex (corner) of the rectangle diagonally (recommended, but optional). 
- Sketch the hyperbola in between the asymptotes.

Example 3

Graph. 
$4 x^{2} - 9 y^{2} - 40 x - 72 y = 8$

$4 x^{2} - 40 x - 9 y^{2} - 72 y = 8 (4 x^{2} - 40 x) + (- 9 y^{2} - 72 y) = 8 4 (x^{2} - 10 x + {(- \frac{10}{2})}^{2}) - 9 (y^{2} + 8 y + {(\frac{8}{2})}^{2}) = 8 + 4 {(- \frac{10}{2})}^{2} - 9 {(\frac{8}{2})}^{2} 4 {(x - 5)}^{2} - 9 {(y + 4)}^{2} = 8 + 4 (25) - 9 (16) 4 {(x - 5)}^{2} - 9 {(y + 4)}^{2} = - 36 \frac{4 {(x - 5)}^{2}}{- 36} + \frac{- 9 {(y + 4)}^{2}}{- 36} = \frac{- 36}{- 36} \frac{{(x - 5)}^{2}}{- 9} + \frac{{(y + 4)}^{2}}{4} = 1 \frac{{(y + 4)}^{2}}{4} - \frac{{(x - 5)}^{2}}{9} = 1$

$Center (5, - 4) a = 3, b = 2 m = \pm \frac{2}{3}$

Example 4

Graph the hyperbola with horizontal branches using equations of the asymptotes:

$y = \pm \frac{5}{4} (x + 4)$

$Center (- 4, 0), horizontal m = \pm \frac{b}{a} = \pm \frac{5}{4} b = 5, b^{2} = 25 a = 4, a^{2} = 16$