Problem 10 - Algebra and Trigonometry

Question

find the standard foTherefore, the standard form of the equation of the parabola that satisfies the given conditions is \(\frac{x^{2}}{16}-\frac{y^{2}}{64}=1\)rm of the equation of each hyperbola satisfying the given conditions. Endpoints of transverse axis: \((-4,0),(4,0) ;\) asymptote: \(y=2 x\)

Step-by-Step Solution

Answer

Therefore, the standard form of the equation of the hyperbola that satisfies the given conditions is \(\frac{x^{2}}{16}-\frac{y^{2}}{64}=1\).

Step 1: Determine the center and the transverse axis

The center of the hyperbola is midway between the two given endpoints of the transverse axis. Since the endpoints are (-4,0) and (4,0), the center is (0,0). The length of the transverse axis is the distance between the two given endpoints, which equals 8 units here.

Step 2: Calculate the semi transverse axis and semi conjugate axis

The length of semi transverse axis 'a' is half the length of the transverse axis. In this case, a = 8/2 = 4. Now, The equation of the asymptote is \(y=2x\), so from it we determine the slope equals 2 which equals b/a (semi conjugate axis/semi transverse axis). Solving for b, we have: b = 2 * a = 2 * 4 = 8.

Step 3: Form the standard equation for the hyperbola

The standard form of the equation of a horizontal hyperbola centered at the origin, with transverse axis along x-axis is \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\). Substituting the found values a = 4 and b = 8, we get: \(\frac{x^{2}}{4^{2}} - \frac{y^{2}}{8^{2}} = 1\) or \(\frac{x^{2}}{16} - \frac{y^{2}}{64} = 1\).

Key Concepts

center of hyperbola

Understanding the center of a hyperbola is crucial as it helps in plotting the shape accurately on a graph. The center of a hyperbola is the point around which the hyperbola is symmetric. It is located at the midpoint of the transverse axis.
In the provided exercise, the endpoints of the transverse axis are (-4, 0) and (4, 0). To find the center, calculate the midpoint of these two points.

• For a midpoint between points ext{(x1, y1)} and ext{(x2, y2)}, use the formula: ext{Midpoint} = ext{((x1 + x2)/2, (y1 + y2)/2)}
• Computing this here gives us ((−4+4)/2, (0+0)/2) = (0, 0). Therefore, the center of the hyperbola is at (0, 0).

Determining the center correctly sets up for solving further attributes of the hyperbola, such as shifts or rotations, if any exist.

transverse axis

The transverse axis is the line segment that runs through the center of the hyperbola and connects the vertices of the hyperbola.
This axis is important as it indicates the direction in which the hyperbola opens. For a horizontal hyperbola, the transverse axis lies along the x-axis, and for a vertical hyperbola, it lies along the y-axis.
In this problem, the transverse axis endpoints are (-4,0) and (4,0), indicating a horizontal hyperbola.

• The length of the transverse axis is the distance between these endpoints: (\(|4 - (-4)| = 8\)).
• Since the transverse axis runs through the center of the hyperbola, it matches the direction in which the hyperbola expands as it diverges. Knowing this helps in determining the placement and orientation of the hyperbola.

By understanding the transverse axis, you can easily determine any major axis lengths and help plot more accurate graphs of the hyperbola.

asymptotes of hyperbola

In the context of hyperbolas, asymptotes serve as guidelines for how the curves of the hyperbola behave at infinity. They are straight lines that the hyperbola approaches but never meets.
The equations of the asymptotes depend on the orientation of the hyperbola.
For a hyperbola centered at the origin with a horizontal transverse axis, the asymptotes are typically of the form: \(y = \pm \frac{b}{a}x\) where \(a\) is the semi-transverse axis and \(b\) is the semi-conjugate axis.

• From the exercise, we have \(y = 2x\) as one of the asymptotes.
• Given \(b/a = 2\), we substitute \(a = 4\) to find \(b\).
• This calculation results in \(b = 2 \times 4 = 8.\)

Asymptotes help guide the drawing of the hyperbola's shape and provide insight into its spread and direction. They are essential for complete descriptions of the hyperbola.

standard form of hyperbola

The standard form of a hyperbola's equation depends on its orientation. For a hyperbola with a horizontal transverse axis centered at the origin, the equation is:\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] where \(a\) is the length of the semi-transverse axis and \(b\) is the semi-conjugate axis.
For vertical hyperbolas, the standard form is:\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \].

• From the exercise, we find \(a = 4\) and \(b = 8\).
• Substituting these into the horizontal form gives us:\[ \frac{x^2}{16} - \frac{y^2}{64} = 1. \]

This equation represents the hyperbola in question. Always ensure you determine \(a\) and \(b\) correctly for an accurate representation of the hyperbola. Knowing the correct form makes it easier to further analyze the properties of the hyperbola, such as its asymptotes and vertices.