Answer to Question #146744 in Analytic Geometry for Kookie

Question

An engineer is working on making a diagram for the cooling tower he was going to to construct in the future. Cooling towers can be seen as towers with-inward curving sides, or simply, hyper-bola looking sides. His daughter ask him how high the center of the hyper-bolic sides of the tower was Y=+- 24/7x + 15, and he answers that the asymtotes of the hyperbola are the following:, and the area of its auxiliary rectangle os 168 square units. The daughter then finds the foci of the hyperbola. At what points on the cartesian plane are these foci found on?

Answer 1

$\text{Let } F_1\text{ and } F_2 \text{ be foci of a hyperbola and let M be}\\ \text{an arbitrary point of a hyperbola. Then we have:}\\ |MF_2|-|MF_1|=2a\text{ where } a=\text{const}.\\ |F_1F_2|=2c\text{ where } c=\text{const}.\\ \text{We denote }c^2-a^2=b^2.\\ \text{Then we have:}\\ \begin{cases} 2b\cdot2a=168,\\ \frac{b}{a}=\frac{24}{7}. \end{cases}\\ a=\frac{7}{24}b\\ 2b\cdot2\frac{7}{24}b=168\\ \frac{7b^2}{6}=168\\ b^2=\frac{6\cdot 168}{7}\\ b^2=144\\ a^2=(\frac{7}{24})^2b^2\\ a^2=(\frac{7}{24})^2\frac{6\cdot 168}{7}\\ a^2=12.25\\ c^2=a^2+b^2\\ c^2=12.25+144\\ c^2=156.25\\ c=12.5\\ \text{Then } F_1(12.5,15) \text{ and } F_2(-12.5,15).$