Final answer:
The equation of the hyperbola with a rectangle with the given vertices and with diagonals as the asymptotes is №(x - 1)^2/36 - (y + 1)^2/16 = 1.
Step-by-step explanation:
To find the equation of the hyperbola when given four vertices of a rectangle and knowing the diagonals of the rectangle are the asymptotes, we can follow these steps:
- Identify the center of the rectangle which is also the center of the hyperbola. The center is the midpoint of the diagonals of the rectangle and can be found using the midpoint formula.
- Determine the lengths of the transverse and conjugate axes. The lengths are twice the distances from the center to the vertices along the axes.
- Calculate the slopes of the diagonals to identify the slopes of the asymptotes. Since the diagonals of the rectangle are the asymptotes for the hyperbola, use the coordinates of the vertices to find the slopes.
- With the center (h, k), the lengths of the transverse axis (2a) and conjugate axis (2b), and the slopes of the asymptotes, write the equation of the hyperbola. If the transverse axis is parallel to the x-axis, the equation has the form №(x - h)^2/a^2 - (y - k)^2/b^2 = 1, where a is half the length of the transverse axis, and b is half the length of the conjugate axis.
In this case, the center of the rectangle (and therefore the hyperbola) is at the midpoint of the diagonals with vertices (-5,3) and (7,-5). The midpoint is found to be (1,-1). The length of the transverse axis is 12 (from -5 to 7 along the x-axis) and the length of the conjugate axis is 8 (from 3 to -5 along the y-axis). This means a=6 and b=4. The slopes of the diagonals are found to be ±(8/12) = ±(2/3), which represent the slopes of the asymptotes, indicating a horizontal transverse axis.
Therefore, the equation of the hyperbola is:
№(x - 1)^2/6^2 - (y + 1)^2/4^2 = 1
or in its simplified form:
№(x - 1)^2/36 - (y + 1)^2/16 = 1.