Question

Find the equation of the given central conic. Hyperbola with a vertex at \((0,-3)\) and eccentricity \(\frac{3}{2}\)

Short answer

The equation of the hyperbola is \(\frac{(y+3)^2}{9} - \frac{4x^2}{45} = 1\).

Step-by-step solution

Identify the standard form of hyperbola equation

The standard form of the equation of a hyperbola with a vertical transverse axis is \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \), where \((h, k)\) is the center.

Analyze given vertex and find center

The vertex is at \((0, -3)\), suggesting the hyperbola's transverse axis is vertical and its center is also at \((0, -3)\). Thus, \(h = 0\) and \(k = -3\).

Determine eccentricity components

For a hyperbola, the eccentricity \(e\) is given by \(e = \frac{c}{a}\). The problem states that \(e = \frac{3}{2}\), so we have \(c = \frac{3}{2}a\).

Use vertex information to find \(a\)

The vertex form citation gives us \((0, k-a) = (0, -3)\). Since our center is \((0, -3)\), \(k-a = -3\). This implies \(a = 3\).

Calculate \(c\) and \(b\) values

Using \(c = \frac{3}{2}a\) where \(a = 3\), we have \(c = \frac{3}{2} \times 3 = \frac{9}{2}\). Using the relationship \(c^2 = a^2 + b^2\), substitute \(c = \frac{9}{2}\) and \(a = 3\): \(\left(\frac{9}{2}\right)^2 = 3^2 + b^2\).

Solve for \(b^2\)

Calculating \(\frac{81}{4} = 9 + b^2\). Solving for \(b^2\) gives: \(b^2 = \frac{81}{4} - \frac{36}{4} = \frac{45}{4}\).

Write the equation of the hyperbola

Substitute the values into the standart form: \(\frac{(y+3)^2}{3^2} - \frac{x^2}{\frac{45}{4}} = 1\), which simplifies to \(\frac{(y+3)^2}{9} - \frac{4x^2}{45} = 1\).

Key concepts

Eccentricity

Eccentricity is a measure of how much a conic section deviates from being circular. For a hyperbola, the eccentricity always exceeds one. It is denoted by the symbol \(e\).

• An eccentricity of 1 results in a parabola.
• For circles, \(e = 0\).
• Ellipses have an eccentricity between 0 and 1.

In the context of hyperbolas, eccentricity connects directly to the shape by determining how "open" it is. The formula for eccentricity in hyperbolas is \(e = \frac{c}{a}\), where \(c\) is the distance from the center to a focus and \(a\) is the distance from the center to a vertex.In our particular exercise, the problem states that \(e = \frac{3}{2}\), meaning the hyperbola is relatively open. This specific value of \(e\) helped identify the distance \(c\) because it allowed us to express it in terms of \(a\) as \(c = \frac{3}{2}a\). This is a critical step in crafting the equation of the hyperbola.

Vertex of Hyperbola

A vertex is a critical defining point of a hyperbola. It is one of the two farthest points on the hyperbola's transverse axis from the center. The position of the vertices helps in determining the overall size and shape of the hyperbola. The transverse axis is the line segment that passes through the vertices and the center. In this exercise, the given vertex is at \((0, -3)\). Knowing the vertex provided crucial insight: it suggested that the transverse axis is vertical, and the center could also be at \((0, -3)\), inline with the vertex itself. Vertices are used in the standard form to determine \(a\), the distance from the center to a vertex. As per the problem's context, identifying the position of the vertex and recognizing it helps determine how "stretched" or "compressed" the hyperbola will appear on the graph.

Transverse Axis

The transverse axis of a hyperbola holds mathematical significance similar to that of the major axis in ellipses. It intersects the hyperbola at its vertices and is intrinsic to defining the shape's orientation and dimensions.

• For a vertical transverse axis, the equation of the hyperbola is given by: \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)
• If the transverse axis were horizontal, the equation would swap variables, yielding: \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\)

In this exercise, the center was identified based on the vertex location. With a vertex at \((0, -3)\), the problem indicates that the transverse axis must be vertical. This fact directed the equation towards using the appropriate standard form.

Standard Form of Hyperbola

The standard form of the equation is pivotal in defining a hyperbola's characteristics and graphing it accurately. It allows for a clear representation of the hyperbola's orientation, size, and position. The general standard form of a hyperbola depends on whether it has a vertical or horizontal transverse axis. - A hyperbola with a vertical transverse axis takes the form: \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \).- For a horizontal transverse axis, the form becomes: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \).In using the standard form, elements like \(h\), \(k\), \(a\), and \(b\) are not just abstract variables but carry spatial meaning:

• \((h, k)\): the center of the hyperbola.
• \(a\): the distance from the center to each vertex along the transverse axis.
• \(b\) relates to the distance perpendicular to the transverse axis.

In the scenario from the exercise, determining \(a^2 = 9\), \(b^2 = \frac{45}{4}\), \(h = 0\), and \(k = -3\) lets us precisely form the equation: \(\frac{(y+3)^2}{9} - \frac{4x^2}{45} = 1\). This equation is the blueprint of the hyperbola with all necessary parameters identified.