https://file.aiquickdraw.com/mathbot/file/1727240096641wnaj6sig.jpg

The equation in the image is the hyperbola equation:

\[
\frac{x^2}{16} - \frac{y^2}{52} = 1
\]

This is the standard form of a hyperbola, which is written as:

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

Here, we can compare and identify the constants:
- \( a^2 = 16 \), so \( a = 4 \)
- \( b^2 = 52 \), so \( b \approx 7.21 \) (approximately the square root of 52)

This hyperbola is oriented horizontally since the \( x^2 \)-term comes first and is positive. The center is at the origin \( (0, 0) \), and the transverse axis lies along the x-axis.

### Key features of this hyperbola:
1. **Vertices**: The vertices are at \( (\pm a, 0) = (\pm 4, 0) \).
2. **Foci**: The foci are located at \( (\pm c, 0) \), where \( c = \sqrt{a^2 + b^2} \). Calculating \( c \):

\[
c = \sqrt{16 + 52} = \sqrt{68} \approx 8.25
\]

So, the foci are at \( (\pm 8.25, 0) \).
3. **Asymptotes**: The asymptotes are the lines through the center with slopes \( \pm \frac{b}{a} = \pm \frac{7.21}{4} \approx \pm 1.8 \). Thus, the equations of the asymptotes are:

\[
y = \pm 1.8x
\]

If you're asked to select the graph of this hyperbola, you should look for a graph that has:
- Two branches opening left and right (since it's horizontal),
- Vertices at \( (-4, 0) \) and \( (4, 0) \),
- Asymptotes with the slope of approximately 1.8.

Would you like more details or explanations? 

Here are some related questions for further practice:
1. What changes if the equation was \( \frac{y^2}{16} - \frac{x^2}{52} = 1 \)?
2. How do you find the equation of the asymptotes for a hyperbola?
3. What is the effect of changing \( a \) or \( b \) in the hyperbola equation?
4. How would you graph this hyperbola by hand?
5. What would the equation look like if the hyperbola were vertical instead of horizontal?

**Tip**: Always identify whether a hyperbola opens horizontally or vertically by checking which variable has the positive term.

Which is the graph of the hyperbola \( \frac{x^2}{16} - \frac{y^2}{52} = 1 \)?

Learn how to graph the hyperbola \( \frac{x^2}{16} - \frac{y^2}{52} = 1 \), including finding the vertices, foci, and asymptotes. This problem covers key concepts of hyperbolas and is suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation