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Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. $$4 y^{2}-x^{2}=1$$

Short Answer

Expert verified
Vertices: (0, ±1/2); Foci: (0, ±√5/2); Asymptotes: y = ±1/2x.

Step by step solution

01

Rewrite the Equation in Standard Form

The given equation is \(4y^2 - x^2 = 1\). Begin by rewriting it in the standard form of a hyperbola, which is \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). Divide every term by 1 to maintain equality:\[\frac{4y^2}{1} - \frac{x^2}{1} = \frac{1}{1}\]This simplifies to \(\frac{y^2}{\frac{1}{4}} - \frac{x^2}{1} = 1\), so \(a^2 = \frac{1}{4}\) and \(b^2 = 1\).
02

Identify the Vertices

Since the hyperbola is of the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the vertices are located at \((0, \pm a)\). Calculate \(a = \sqrt{\frac{1}{4}} = \frac{1}{2}\). Thus, the vertices are \((0, \frac{1}{2})\) and \((0, -\frac{1}{2})\).
03

Find the Foci

The foci of the hyperbola are located at \((0, \pm c)\) for a vertical hyperbola, where \(c^2 = a^2 + b^2\). Calculate \(c\):\[c^2 = \frac{1}{4} + 1 = \frac{5}{4}\]\(c = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}\).Therefore, the foci are \((0, \frac{\sqrt{5}}{2})\) and \((0, -\frac{\sqrt{5}}{2})\).
04

Determine the Asymptotes

For a vertical hyperbola, the asymptotes are given by the equations \(y = \pm \frac{a}{b}x\). Calculate \(\frac{a}{b}\):\[\frac{\frac{1}{2}}{1} = \frac{1}{2}\]Thus, the asymptotes are \(y = \frac{1}{2}x\) and \(y = -\frac{1}{2}x\).
05

Sketch the Graph

To sketch the hyperbola, draw the asymptotes as lines that cross the y-axis at the origin and have slopes of \(\frac{1}{2}\) and \(-\frac{1}{2}\). Mark the vertices and foci on the y-axis. Use the asymptotes as guides to draw the hyperbola opening upwards and downwards from the vertices.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Hyperbola Vertices
In the realm of hyperbolas, vertices play a crucial role. They mark the points where the hyperbola intersects its axis of symmetry. For a vertical hyperbola, which revolves around the y-axis, these vertices lie on the y-axis.
Begin by identifying the standard form of a hyperbola equation:
  • When in the form of \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the vertices are located at \((0, \pm a)\).
In our example, the solution found that \(a^2 = \frac{1}{4}\), which means \(a = \frac{1}{2}\). Therefore, the vertices are located at \((0, \frac{1}{2})\) and \((0, -\frac{1}{2})\). These points are the closest that the branches of the hyperbola come to one another on the y-axis.
Remember, vertices help sketch the hyperbola by pinpointing its narrowest points. These act as anchors, ensuring that the branches open symmetrically. Keep this in mind as you work with hyperbolas, as it's a consistent pattern across different forms.
Discovering the Foci of a Hyperbola
The foci of a hyperbola are another fundamental aspect, shaping the unique curves of the hyperbola's branches. For hyperbolas, these foci are found inside the arms of the curves, located farther from the center compared to the vertices.
The formula for the foci requires calculations of \(c\), which is found using the equation:
  • \(c^2 = a^2 + b^2\)
In our equation, \(a^2 = \frac{1}{4}\) and \(b^2 = 1\), resulting in \(c^2 = \frac{1}{4} + 1 = \frac{5}{4}\). Solving for \(c\) gives \(c = \frac{\sqrt{5}}{2}\). So, the foci are located at \((0, \frac{\sqrt{5}}{2})\) and \((0, -\frac{\sqrt{5}}{2})\).
Understanding the position of the foci is vital for constructing the hyperbola. They lie along the y-axis and indicate where the hyperbola extends and contracts. While the vertices show us the narrowest points, the foci tell us how far the hyperbola stretches.
The Role of Asymptotes in Hyperbolas
Asymptotes in hyperbolas are invisible guideposts that help shape the graph's overall form. Each arm of the hyperbola approaches its corresponding asymptote but never touches it. Finding these lines provides a skeleton around which the hyperbola forms.
For vertical hyperbolas, the asymptote equations are:
  • \(y = \pm \frac{a}{b}x\)
In this scenario, with \(a = \frac{1}{2}\) and \(b = 1\), the asymptotes are defined by the slopes \(y = \frac{1}{2}x\) and \(y = -\frac{1}{2}x\).
These lines cross the y-axis at the origin, serving as a crucial visual aid when sketching. Knowing the angles at which the hyperbola arms open—indicated by the slopes—ensures accurate depiction.
Asymptotes reinforce the symmetrical character of the hyperbola, ensuring balance and proper direction of its branches. When dealing with these components together, the hyperbola becomes much easier to understand and graph.

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Most popular questions from this chapter

Determine the \(X Y\) -coordinates of the given point if the coordinate axes are rotated through the indicated angle. $$(\sqrt{2}, 4 \sqrt{2}), \quad \phi=45^{\circ}$$

Sketch the curve given by the parametric equations. $$x=\sin t, \quad y=\sin 2 t$$

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In this section we stated that parametric equations contain more information than just the shape of a curve. Write a short paragraph explaining this statement. Use the following example and your answers to parts (a) and (b) below in your explanation. The position of a particle is given by the parametric equations $$x=\sin t \quad y=\cos t$$ where \(t\) represents time. We know that the shape of the path of the particle is a circle. (a) How long does it take the particle to go once around the circle? Find parametric equations if the particle moves twice as fast around the circle. (b) Does the particle travel clockwise or counterclockwise around the circle? Find parametric equations if the particle moves in the opposite direction around the circle.

Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center, foci, vertices, and lengths of the major and minor axes. If it is a parabola, find the vertex, focus, and directrix. If it is a hyperbola, find the center, foci, vertices, and asymptotes. Then sketch the graph of the equation. If the equation has no graph, explain why. $$2 x^{2}+y^{2}=2 y+1$$

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