Find an equation for the hyperbola described. Graph the equation by hand. Center at \( (0,0) \); focus at \( (0,-10) \); vertex at \( (0,6) \)

To find the equation of the hyperbola, we can use the standard form for a hyperbola with its center at the origin: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] where \( a \) is the distance from the center to the vertex, and \( b \) is the distance from the center to the focus. Given that the center is at \( (0,0) \), the vertex is at \( (0,6) \), and the focus is at \( (0,-10) \), we can determine the values of \( a \) and \( b \) as follows: 1. The distance from the center to the vertex is \( a \), so \( a = 6 \). 2. The distance from the center to the focus is \( c \), where \( c = 10 \) because the focus is 10 units below the center. For a hyperbola, the relationship between \( a \), \( b \), and \( c \) is given by: \[ c^2 = a^2 + b^2 \] We can solve for \( b^2 \): \[ b^2 = c^2 - a^2 \] Substituting the known values: \[ b^2 = 10^2 - 6^2 \] \[ b^2 = 100 - 36 \] \[ b^2 = 64 \] Now we have \( a^2 = 36 \) and \( b^2 = 64 \). The equation of the hyperbola is: \[ \frac{x^2}{36} - \frac{y^2}{64} = 1 \] To graph this hyperbola by hand, follow these steps: 1. Plot the center of the hyperbola at \( (0,0) \). 2. Plot the vertices, which are \( (0,6) \) and \( (0,-6) \). 3. Plot the foci, which are \( (0,-10) \) and \( (0,10) \). 4. Draw the asymptotes of the hyperbola, which are the lines \( y = \pm \frac{b}{a}x \). In this case, the asymptotes are \( y = \pm \frac{8}{6}x \) or \( y = \pm \frac{4}{3}x \). 5. Sketch the hyperbola, ensuring it passes through the vertices and foci, and is symmetric with respect to the asymptotes. Here's a rough sketch of the hyperbola: ``` y | | * | / \ | / \ | / \ | / \ | / \ | / \ |/ \ +----------------+ -10 10 ``` The hyperbola opens up and down, and the branches are closer together near the center and get further apart as they extend away from the center.