Problem 78 - College Algebra

Question

Graph \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) and \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=-1\) in the same viewing rectangle for values of \(a^{2}\) and \(b^{2}\) of your choice. Describe the relationship between the two graphs.

Step-by-Step Solution

Answer

The hyperbola given by \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) is horizontal whereas the hyperbola given by \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=-1\) is vertical. This orientation is determined by the positive or negative sign of the constant on the right side of the equation.

Step 1: Choosing Values of \(a^2\) and \(b^2\)

Choose specific values for \(a^2\) and \(b^2\). Let's choose \(a^2=4\) and \(b^2=1\). We substitute these values into both equations: \(\frac{x^{2}}{4}-\frac{y^{2}}{1}=1\) and \(\frac{x^{2}}{4}-\frac{y^{2}}{1}=-1\)

Step 2: Graphing the First Equation

Graph \(\frac{x^{2}}{4}-\frac{y^{2}}{1}=1\). On the x-axis, points will be at \(x=\pm2\) and on y-axis, it will be \(y=\pm i\). It will be a hyperbola opening left and right with vertices at \(x=\pm2\).

Step 3: Graphing the Second Equation

Graph \(\frac{x^{2}}{4}-\frac{y^{2}}{1}=-1\). On the x-axis, points will be at \(x=\pm2i\) and on y-axis, it will be \(y=\pm1\). It will be a hyperbola opening up and down with vertices at \(y=\pm1\).

Step 4: Interpreting the Relationship

Compare the two hyperbolas and try to identify a relationship. Since the only difference in equations was the sign, changing it changed our orientation from horizontal to vertical. The hyperbola from equation one opens horizontally while the hyperbola from equation two opens vertically.