Problem 44 - Precalculus with Calculus Previews

Question

In Problems \(21-44,\) find an equation of the hyperbola that satisfies the given conditions. Eccentricity \(\sqrt{10},\) endpoints of conjugate axis (-5,4),(-5,10)

Step-by-Step Solution

Answer

The hyperbola equation is \(\frac{(x + 5)^2}{1} - \frac{(y - 7)^2}{9} = 1\).

Step 1: Identify the Center

The given conjugate axis endpoints are (-5,4) and (-5,10). Find the midpoint of this line as it represents the center of the hyperbola. Using the midpoint formula: \[\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\]we get the center as \((−5, 7)\).

Step 2: Find the Length of Conjugate Axis

The length of the conjugate axis is found by calculating the distance between the endpoints \((−5, 4)\) and \((−5, 10)\) using the distance formula:\[\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]Substituting the values yields \(|10 - 4| = 6\). Thus, the length of the conjugate axis is 6.

Step 3: Determine b and Coordinate Axes

The length of the conjugate axis \((2b)\) is 6, thus \(b = 3\). Since the endpoints of the conjugate axis are vertical, this indicates the transverse axis is horizontal, centered at \((x+5)^2/(a^2) - (y-7)^2/(b^2) = 1\).

Step 4: Solve for a

We know the eccentricity is \(e = \sqrt{10}\), and for hyperbolas \(e = \sqrt{1 + (b^2/a^2)}\). Substituting \(b = 3\) gives:\[\sqrt{10} = \sqrt{1 + \frac{9}{a^2}}\]Squaring both sides and solving for \(a^2\):\[10 = 1 + \frac{9}{a^2}\]\[9 = 10a^2 - 1a^2\)\(a = 1\].

Step 5: Write the Equation of the Hyperbola

With \(a = 1\), \(b = 3\), and center \((-5, 7)\), the equation for a hyperbola with a horizontal transverse axis is:\[\frac{(x + 5)^2}{a^2} - \frac{(y - 7)^2}{b^2} = 1\]Substituting the values derived gives the equation:\[\frac{(x + 5)^2}{1} - \frac{(y - 7)^2}{9} = 1\].

Key Concepts

Eccentricity

Eccentricity is a core characteristic that distinguishes hyperbolas from other conic sections like ellipses and parabolas. It's a measure of how stretched or elongated the hyperbola is. In mathematical terms, the eccentricity (\(e\)) of a hyperbola is defined as \(e = \sqrt{1 + \frac{b^2}{a^2}}\), where \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively. For hyperbolas, the eccentricity is always greater than 1, indicating they never form closed curves.
In this exercise, we are given an eccentricity of \(\sqrt{10}\), suggesting that this hyperbola is significantly stretched. This is directly linked to how far apart the branches of the hyperbola are, and assists in determining the precise geometric shape for forming the equation of the hyperbola.

Conjugate Axis

The conjugate axis is an essential component in defining the structure of a hyperbola. It is perpendicular to the transverse axis and, unlike the transverse axis, does not contain any real points of intersection with the hyperbola itself. Instead, it helps inform the size and shape of the hyperbola.The hyperbola described in the problem exhibits a conjugate axis with endpoints at (-5, 4) and (-5, 10), vertically aligned. The length of this axis is crucial as it informs the calculations for \(b\)—the semi-minor axis. In this case, the length of 6 divides into \(2b\), resulting in \(b = 3\). This helps us understand that the size of the conjugate axis directly influences the equation of the hyperbola.

Midpoint Formula

The midpoint formula is a simple yet powerful tool in geometry, crucial for identifying specific points like the center of a hyperbola or other geometric figures. Given two endpoints (\((x_1, y_1)\) and \((x_2, y_2)\)), the midpoint is calculated by \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\).
In this exercise, the endpoints of the conjugate axis are given as (-5, 4) and (-5, 10). Using the midpoint formula, the center of the hyperbola is found to be (-5, 7). Recognizing and accurately determining the center is fundamental for deriving the full equation of a hyperbola, as it serves as the position around which all other measurements are oriented.

Distance Formula

The distance formula is employed to compute the distance between two points in a coordinate plane. Given endpoints (\((x_1, y_1)\) and \((x_2, y_2)\)), it's expressed as \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
In this scenario, to ascertain the length of the conjugate axis of the hyperbola, this formula is vital. Applying it to the endpoints (-5, 4) and (-5, 10) yields a distance of 6. Calculating this distance accurately leads to determining \(b\), which then feeds into formulating the general equation of the hyperbola. This step ensures the hyperbola is precisely defined according to the problem's specific parameters.