Precalculus Examples

$\frac{{(x + 3)}^{2}}{16} + \frac{{(y + 1)}^{2}}{4} = 1$

Step 1

Simplify each term in the equation in order to set the right side equal to $1$ . The standard form of an ellipse or hyperbola requires the right side of the equation be $1$ .

$\frac{{(x + 3)}^{2}}{16} + \frac{{(y + 1)}^{2}}{4} = 1$