Math / Geometry

QuestionSketch a graph of $\frac{x^{2}}{16}-\frac{y^{2}}{25}=1$
Clear All $\square$ Draw: $\square$
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Studdy Solution

STEP 1

1. The equation given is in the form of a hyperbola.
2. The hyperbola is centered at the origin.
3. The transverse axis is along the x-axis since the x-term is positive.

STEP 2

1. Identify the type of conic section and its orientation.
2. Determine the vertices and asymptotes.
3. Sketch the hyperbola using the identified features.

STEP 3

Identify the type of conic section and its orientation:
The equation $\frac{x^{2}}{16} - \frac{y^{2}}{25} = 1$ is a hyperbola. Since the x-term is positive, the hyperbola opens left and right along the x-axis.

STEP 4

Determine the vertices and asymptotes:
- The standard form of a hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ . - Here, $a^{2} = 16$ and $b^{2} = 25$ . - Therefore, $a = 4$ and $b = 5$ .
Vertices: - The vertices are at $(\pm a, 0) = (\pm 4, 0)$ .
Asymptotes: - The equations of the asymptotes are $y = \pm \frac{b}{a}x = \pm \frac{5}{4}x$ .

STEP 5

Sketch the hyperbola using the identified features:
1. Draw the coordinate axes.
2. Plot the vertices at $(4, 0)$ and $(-4, 0)$ .
3. Draw the asymptotes as dashed lines through the origin with slopes $\pm \frac{5}{4}$ .
4. Sketch the branches of the hyperbola opening left and right, approaching the asymptotes.

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