Question

For the following exercises, find the Cartesian equation describing the given shapes. A hyperbola with vertices at \((3,-2)\) and \((-5,-2)\) and foci at \((-2,-6)\) and \((-2,4)\)

Short answer

Equation: \(\frac{(x+1)^2}{16} - \frac{(y+2)^2}{9} = 1\).

Step-by-step solution

Identify the Hyperbola Type

Since the vertices are on a horizontal line (the same y-coordinate), this indicates that the transverse axis is horizontal. The hyperbola is of the form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \).

Calculate the Center

The center of a hyperbola is the midpoint between the vertices. The midpoint of the vertices \((3, -2)\) and \((-5, -2)\) is \((-1, -2)\). So, the center \((h, k)\) is \((-1, -2)\).

Find the Value of \(a\)

The distance between the vertices is 8 units \((3 - (-5)) = 8\). The distance from the center to each vertex \(a\) is half of this, so \(a = 4\).

Calculate the Value of \(c\)

The distance between the center \((-1, -2)\) and each focus \((-2, -6)\) and \((-2, 4)\) is 5 units. This is the value \(c\), so \(c = 5\).

Calculate \(b\) Using the Relationship \(c^2 = a^2 + b^2\)

Using the relationship \(c^2 = a^2 + b^2\), we have \(5^2 = 4^2 + b^2\). Hence, \(25 = 16 + b^2\). Therefore, \(b^2 = 9\), hence \(b = 3\).

Write the Cartesian Equation

Substitute \(h = -1\), \(k = -2\), \(a = 4\), and \(b = 3\) into the hyperbola equation: \(\frac{(x+1)^2}{16} - \frac{(y+2)^2}{9} = 1\).

Key concepts

Cartesian equation

The Cartesian equation is essential for describing the geometric properties of a hyperbola. In this context, the term "Cartesian equation" refers to the algebraic formula that represents the hyperbola's shape on a coordinate plane. In general, a hyperbola's equation looks like this:

• For a horizontal transverse axis: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)
• For a vertical transverse axis: \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)

To identify which equation to use, check the orientation of the hyperbola. For the given exercise, since the vertices share the same y-coordinate, it's clear the transverse axis is horizontal, indicating the use of the first equation. The terms \((h, k)\) represent the center of the hyperbola, \(a\) is the distance from the center to each vertex along the horizontal axis, and \(b\) relates to the distance along the vertical axis.

vertices of hyperbola

The vertices of a hyperbola are two crucial points that define the width of the hyperbola. In the Cartesian plane, they lie along the transverse axis. In the exercise given, the vertices are located at \((3, -2)\) and \((-5, -2)\). This means both vertices share the same y-coordinate.To find the center of the hyperbola, we calculate the midpoint of these vertices. The midpoint formula \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \) gives us the center at \((-1, -2)\). This midpoint helps in understanding the symmetry of the hyperbola. The distance between these vertices, in this case, is 8 units (the difference between the x-coordinates), which is divided evenly on either side of the center, giving \(a = 4\). This value \(a\) is crucial for constructing the Cartesian equation of the hyperbola.

transverse axis

The transverse axis is a fundamental concept that guides the form and orientation of a hyperbola. Essentially, it is the axis that passes through the vertices and represents the hyperbola's longest diameter.In standard situations, the transverse axis can be horizontal or vertical, determining the direction the hyperbola opens. In our problem, due to the vertices having the same y-coordinate, the axis is horizontal. This horizontal orientation indicates that the hyperbola stretches longest horizontally.The length of the transverse axis is twice the distance from the center to either vertex. Here, since \(a = 4\), the total length becomes 8 units. This axis not only dictates the hyperbola's form but also helps to determine which standard equation form should be applied, shaping the hyperbola's graphical representation.

foci of hyperbola

The foci of a hyperbola are distinct points that further define the shape and properties of the curve. Unlike the vertices, which lie directly on the curve, the foci are internal and lie on the same axis as the vertices. They are important in understanding the geometric and algebraic properties of hyperbolas.In the exercise provided, the foci are at \((-2, -6)\) and \((-2, 4)\). Notably, they don’t share a y-coordinate with the center or vertices, but they have significance in revealing how the hyperbola is stretched.To determine the distance \(c\) (between the center and a focus), you measure directly from the center to one focus. Here, \(c = 5\). The relationship \(c^2 = a^2 + b^2\) ensures the hyperbola's equation is linked with these foci. With \(b\) calculated, along with \(a\) from the vertices, this relationship helps form the complete Cartesian equation and paints a full picture of the hyperbola's behavior on the plane.