For $\frac{x^{2}}{49}-\frac{y^{2}}{64}=1$
(a) Identify the center.
(b) Identify the vertices.
(c) Identify the foci.
(d) Write equations for the asymptotes.
(e) Graph the hyperbola.
Give exact answers. Express numbers in simplest form.
Part: $0 / 5$
Answer
To graph the hyperbola, plot the center at \((0,0)\), the vertices at \((\pm7, 0)\), and the foci at \((\pm\sqrt{113}, 0)\). Draw the asymptotes with slopes \(\pm8/7\). The hyperbola opens to the left and right because the x term is positive.
Step 1 ：The center of the hyperbola is at the origin \((0,0)\) because there are no shifts in the equation.
Step 2 ：The vertices are found by taking the square root of the denominator under the x term, which gives us \(a = 7\). So, the vertices are at \((\pm7, 0)\).
Step 3 ：The foci are found by calculating \(c = \sqrt{a² + b²} = \sqrt{49 + 64} = \sqrt{113}\). So, the foci are at \((\pm\sqrt{113}, 0)\).
Step 4 ：The equations for the asymptotes are derived from the slopes \(\pm b/a\). Here, \(b = 8\) and \(a = 7\), so the slopes are \(\pm8/7\). The equations of the asymptotes are \(y = \pm(8/7)x\).
Step 5 ：To graph the hyperbola, plot the center at \((0,0)\), the vertices at \((\pm7, 0)\), and the foci at \((\pm\sqrt{113}, 0)\). Draw the asymptotes with slopes \(\pm8/7\). The hyperbola opens to the left and right because the x term is positive.
