Hyperbolas - Example 1: Sketching x^2/16 - y^2/48 = 1

Previously, we've learned how to find all of the features of the standard hyperbola x^2/a^2 - y^2/b^2 = 1. In this video, we use this knowledge to sketch x^2/16 - y^2/48 = 1. Firstly, we convert the equation to standard form: x^2/16 - y^2/48 = x^2/4^2 - y^2/(4√3)^2 = 1 Thus a = 4 and b = 4√3. So the vertices are located at (+/-a, 0) = (+/-4, 0). Since b^2 = c^2 - a^2, we solve c to be c = 8, and thus the foci are located at (+/-c, 0) = (+/-8, 0). And since c = ae, the eccentricity e = 2. The directrices are given by the equation x = +/-a/e = +/-2. Finally the asymptotes are given by the equation y = +/-(b/a)x = +/-√3x. With all of this information, we can sketch the hyperbola. Thanks for watching. Please give me a "thumbs up" if you have found this video helpful. Please ask me a maths question by commenting below and I will try to help you in future videos. Follow me on Twitter! twitter.com/MasterWuMath