Problem 13 - Precalculus with Limits

Question

In Exercises 9-22, find the center, vertices, foci, and the equations of the asymptotes of the hyperbola, and sketch its graph using the asymptotes as an aid. \(\dfrac{y^2}{1}-\dfrac{x^2}{4}=1\)

Step-by-Step Solution

Answer

The center is at the origin (0, 0), vertices are at points (0,-1) and (0,1), the Foci are at points (0, -\(\sqrt{5}\)) and (0, \(\sqrt{5}\)), the equations of the asymptotes are \(y = \pm \frac{x}{2}\), and the graph of the hyperbola opens up and down.

Step 1: Identify the parameters of the hyperbola

The given equation can be written in the standard form of a hyperbola: \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). Comparing the given equation and standard form we conclude: \(a^2 = 1\) and \( b^2 = 4\) which give \( a = 1\) and \(b = 2\). Here, the center of the hyperbola is at the origin (0,0) because there are no constants added or subtracted to x or y.

Step 2: Find vertices and foci of the hyperbola

The vertices are (\(0, \pm a\)) = (0, \(\pm 1\))= (0,-1) and (0,1). The foci can be found by using \( c = \sqrt{a^2 + b^2} = \sqrt{5}\). Since the hyperbola is oriented vertically, the foci are located at (\(0, \pm c\)) = (0, \(\pm \sqrt{5}\)).

Step 3: Find the equations of the asymptotes

For a hyperbola centered at the origin and oriented vertically, the equations of the asymptotes are \( y = \pm \frac{a}{b}*x = \pm \frac{1}{2}*x\). So, the equations are \(y = \pm \frac{x}{2}\)

Step 4: Draw the graph of the hyperbola

Plot the points for the center, vertices, and foci to get the visual representation of the hyperbola. The endpoints of the transverse and conjugate axis help with sketching the hyperbola more accurately. The asymptotes guide how the hyperbola opens. In this case, it opens up and down, along the y-axis.

Key Concepts

Standard Form of a Hyperbola

Understanding the standard form of a hyperbola is crucial in graphing and analyzing its properties. The standard form of a hyperbola centered at the origin \( (h, k) = (0,0)\) with a vertical orientation is \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1, \] while for a horizontal orientation, it is \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.\] Here, \(a\) and \(b\) represent distances from the center to the vertices along the transverse axis and the distances from the center to the endpoints of the conjugate axis, respectively.

When graphing a hyperbola, distinguishing whether \(a^2\) is under \(y^2\) or \(x^2\) tells us about the orientation. If \(a^2\) is under \(y^2\), as in our exercise, the hyperbola opens upwards and downwards. If \(a^2\) is under \(x^2\), the hyperbola would open left and right.

Vertices and Foci of a Hyperbola

The vertices of a hyperbola are points where the hyperbola intersects its transverse axis. In our solved exercise, the vertices of the hyperbola are identified as \( (0, \pm a) \), which translates to \( (0, -1) \) and \( (0, 1) \) given that \( a = 1 \).

The foci of a hyperbola are internal points from which the distance to any point on the hyperbola has a constant difference. To locate the foci, the formula \( c = \sqrt{a^2 + b^2} \) is used. For the given hyperbola, we calculate \( c \) as \( \sqrt{5} \), which results in the foci being at \( (0, \pm \sqrt{5}) \). These points are critical for understanding the hyperbola's shape and direction of opening.

Equations of Asymptotes

Understanding the equations of the asymptotes is essential as they frame the hyperbola and provide a guide for its sketch. Asymptotes are imaginary lines that the hyperbola approaches but never intersects. For a vertically oriented hyperbola with the center at the origin, the equations of the asymptotes are derived from the slope \( \frac{a}{b} \) and are \[ y = \pm \frac{a}{b} \cdot x.\] In the instance of our exercise, this simplifies to \( y = \pm \frac{1}{2} \cdot x \), resulting in two straight lines that pass through the origin and form an 'X' shape. When graphing, these lines help in properly shaping the arms of the hyperbola.

Transverse and Conjugate Axes

The transverse and conjugate axes are fundamental in understanding the orientation and dimensions of a hyperbola. The transverse axis is the line segment that connects the two vertices. It runs along the direction in which the hyperbola opens. The length of the transverse axis is \(2a\), which is the distance between the two vertices.

The conjugate axis is perpendicular to the transverse axis and bisects it at the center. It represents the distance between the 'imaginary vertices' that would be on the hyperbola if it were to continue across the asymptotes. The length of the conjugate axis is \(2b\). These axes lay the foundation for plotting the hyperbola and understanding its reflective symmetry.