Question Find the foci of the hyperbola defined \( \frac{(x-3)^{2}}{16}-\frac{(y+2)^{2}}{49}=1 \). If necessary, ro nearest tenth.

To find the foci of the hyperbola defined by the equation \( \frac{(x-3)^{2}}{16}-\frac{(y+2)^{2}}{49}=1 \), we first identify the center, which is at the point (3, -2). The values of \( a^2 \) and \( b^2 \) can be read directly from the equation: \( a^2 = 16 \) (so \( a = 4 \)) and \( b^2 = 49 \) (thus \( b = 7 \)). Next, we calculate \( c \), where \( c^2 = a^2 + b^2 \). This gives us \( c^2 = 16 + 49 = 65 \), so \( c = \sqrt{65} \approx 8.1 \). The foci are located at \( (h \pm c, k) = (3 \pm 8.1, -2) \), which means the foci are approximately at \( (11.1, -2) \) and \( (-5.1, -2) \). For a touch of historical flair, hyperbolas were studied by ancient Greeks, and they played a crucial role in celestial navigation and predicting orbits of celestial bodies. The term "hyperbola" itself comes from the Greek word "hyperbolē," meaning "excess" or "to throw beyond," highlighting how this curve reaches infinity and defines the boundaries of certain behaviors in mathematics and physics. If you're looking to dive deeper into hyperbolas, consider "Analytic Geometry" by Gordon Fuller for a comprehensive exploration of curves and conics. Another great read is "Conics" by Patrick Suppes, which touches on different conic sections and delves into both their mathematical properties and real-world applications, including navigation and architecture. Happy reading!