The equation provided in the image is:

$\frac{x^2}{16} - \frac{y^2}{4} - 1$

Let me analyze this and determine its classification or simplify it.

First, rewrite the expression in standard form:

$\frac{x^2}{16} - \frac{y^2}{4} - 1 = 0$

Step 1: Combine terms into a single fraction

Rewrite $-1$ with a denominator of $16$: $\frac{x^2}{16} - \frac{y^2}{4} - 1 = \frac{x^2}{16} - \frac{4y^2}{16} - \frac{16}{16} = \frac{x^2 - 4y^2 - 16}{16}.$

Thus, the equation becomes: $\frac{x^2 - 4y^2 - 16}{16} = 0.$

Step 2: Multiply through by 16 to eliminate the denominator

$x^2 - 4y^2 - 16 = 0.$

Step 3: Recognize the form of the equation

This equation, $x^2 - 4y^2 - 16 = 0$, is in the standard form of a hyperbola: $\frac{x^2}{16} - \frac{y^2}{4} = 1.$

Final Decision:

This is the equation of a hyperbola centered at the origin $(0,0)$, with the following properties:

• Horizontal transverse axis: The hyperbola opens left and right.
• Vertices: Located at $(\pm 4, 0)$ (since $\sqrt{16} = 4$).
• Equation of asymptotes: $y = \pm \frac{1}{2}x$ derived from the slopes $\pm \sqrt{\frac{b^2}{a^2}} = \pm \sqrt{\frac{4}{16}} = \pm \frac{1}{2}$.

Would you like further clarification or visualization? Let me know!

Related Questions:

1. What are the equations of the asymptotes for this hyperbola?
2. How can you graph a hyperbola step by step from its equation?
3. How do the values of $a^2$ and $b^2$ affect the shape of a hyperbola?
4. Can the equation of this hyperbola be converted to polar coordinates?
5. What is the general form of a hyperbola, and how does this example fit it?

Tip:

To graph a hyperbola, start by plotting its vertices and asymptotes, then sketch the two curved branches.