Final answer:
The hyperbola modeling the edges of the pot's neck with given asymptotes and a minimum distance from the center has the equation \(\frac{y^2}{625} - \frac{x^2}{59.16} = 1\), where the slopes of the asymptotes provided the necessary information to find the semi-major axis and the closest distance provided the semi-minor axis.
Step-by-step explanation:
Edward is working on a design for a ceramic pot and needs to write the equation for a hyperbola that models the edges of the pot's neck.
The given asymptotes of the hyperbola are y=3.25x and y=−3.25x, which tells us about the slope of these asymptotes. The distance from the center of the neck to the closest point on the hyperbola is 25 millimeters, which corresponds to the hyperbola's semi-minor axis b, since the vertices are on the x-axis and the center of the hyperbola is at the origin in this context.
For a hyperbola with the center at the origin (0,0), the asymptotes as straight lines with slopes m and -m, and a semi-minor axis of b, the general equation is given by \(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\),
where a is the semi-major axis. The slopes of the asymptotes for the hyperbola are also related to a and b by the equation m = \frac{b}{a}.
Since m is 3.25 in this case, and b equals 25, we can solve for a using a = \frac{b}{m}.
We find that a = \frac{25}{3.25} \approx 7.69 millimeters, which is the distance from the center to each vertex along the x-axis.
Therefore, the equation for this hyperbola is \(\frac{y^2}{25^2} - \frac{x^2}{7.69^2} = 1\), or, after squaring the values of a and b, \(\frac{y^2}{625} - \frac{x^2}{59.16} = 1\).