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

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1

In this video, we use this knowledge to sketch x²/16 - y²/48 = 1.

Firstly, we convert the equation to standard form:

\frac{x^2}{16}-\frac{y^2}{48}=\frac{x^2}{4^2}-\frac{y^2}{\left ( 4\sqrt3 \right )^2}=1

Thus a = 4 and b = 4√3.

So the vertices are located at (±a, 0) = (±4, 0).

Since b² = c² - a², 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.