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Chapter 10: Problem 8

Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. $$x^{2}-y^{2}+4=0$$

Short Answer

Expert verified
Vertices: (2,0), (-2,0); Foci: (2鈭�2,0), (-2鈭�2,0); Asymptotes: y = 卤x.

Step by step solution

01

Simplify the Equation

We begin with the equation of the hyperbola: \[ x^2 - y^2 + 4 = 0 \]Add 4 to both sides to simplify it:\[ x^2 - y^2 = -4 \] Divide every term by -4 to match the standard form:\[ \frac{x^2}{4} - \frac{y^2}{4} = 1 \] This represents a hyperbola centered at (0,0) with its transverse axis along the x-axis.
02

Identify the Vertices

From the standard form, \( \frac{x^2}{4} - \frac{y^2}{4} = 1 \), the hyperbola has the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), where \( a^2 = 4 \) and \( b^2 = 4 \). So, \( a = 2 \) and the vertices are located at (+/-a, 0). Thus, the vertices are at (2, 0) and (-2, 0).
03

Determine the Foci

The foci of the hyperbola can be found using the formula \( c^2 = a^2 + b^2 \). Substitute the values \( a^2 = 4 \) and \( b^2 = 4 \):\[ c^2 = 4 + 4 = 8 \] \( c = \sqrt{8} = 2\sqrt{2} \).Thus, the foci are at \( (2\sqrt{2}, 0) \) and \( (-2\sqrt{2}, 0) \).
04

Find the Asymptotes

The asymptotes of the hyperbola are given by the equations\[ y = \pm \frac{b}{a}x \]. Substitute the values \( a = 2 \) and \( b = 2 \): \[ y = \pm \frac{2}{2}x \] \[ y = \pm x \] These are the equations of the asymptotes.
05

Sketch the Hyperbola

To sketch the hyperbola, plot the center at (0,0), the vertices at (2,0) and (-2,0), and the asymptotes as the lines \( y = x \) and \( y = -x \). Draw the hyperbola opening left and right along the x-axis and approaching the asymptotes as it extends.

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Key Concepts

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

Vertices of a Hyperbola
The vertices of a hyperbola are like anchor points that tell us where the hyperbola begins to curve away. In the equation \( \frac{x^2}{4} - \frac{y^2}{4} = 1 \), we can determine the vertices using the variable \(a\), which represents half the distance between the vertices along the transverse axis (the main axis of the hyperbola).
- Here, \( a^2 = 4 \), so \( a = 2 \). - With the center of the hyperbola at (0, 0), which means the figure is symmetric about this point, the vertices are found at positions (+/-2, 0).So, the vertices are (2, 0) and (-2, 0). These tell us the hyperbola opens horizontally along the x-axis.
Foci of a Hyperbola
Foci are crucial points that help to define the shape and orientation of a hyperbola. They lie on the transverse axis, further out than the vertices.Using the formula \( c^2 = a^2 + b^2 \), where \(a^2 = 4\) and \(b^2 = 4\), we can calculate:- \( c^2 = 4 + 4 = 8 \)- \( c = \sqrt{8} = 2\sqrt{2} \)Thus, the hyperbola's foci are at \((2\sqrt{2}, 0)\) and \((-2\sqrt{2}, 0)\).
These positions indicate that the foci are aligned along the x-axis, each positioned beyond a vertex, signaling that the hyperbola extends horizontally through these points.
Asymptotes of a Hyperbola
Asymptotes function like invisible guidelines that indicate the direction a hyperbola will approach as it stretches away from its center. They don't intersect the hyperbola but lead us in understanding its ultimate path.For a centered hyperbola like \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), asymptotes are given by:\[ y = \pm \frac{b}{a} x \]Here, \( a = 2 \) and \( b = 2 \), so the equations for the asymptotes become:- \( y = x \)- \( y = -x \)These straight lines intersect at the origin (0, 0) and stretch diagonally across the plane, forming a cross-like structure. As the hyperbola curves closer, it approaches but never meets these lines.

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Most popular questions from this chapter

If a projectile is fired with an initial speed of \(v_{0} \mathrm{ft} / \mathrm{s}\) at an angle \(\alpha\) above the horizontal, then its position after \(t\) seconds is given by the parametric equations $$x=\left(v_{0} \cos \alpha\right) t \quad y=\left(v_{0} \sin \alpha\right) t-16 t^{2}$$ (where \(x\) and \(y\) are measured in feet). Show that the path of the projectile is a parabola by eliminating the parameter \(t\).

(a) Use rotation of axes to show that the following equation represents a hyperbola: \(7 x^{2}+48 x y-7 y^{2}-200 x-150 y+600=0\) (b) Find the \(X Y\) - and \(x y\) -coordinates of the center, vertices, and foci. (c) Find the equations of the asymptotes in \(X Y\) - and \(x y\) -coordinates.

A pair of parametric equations is given. (a) Sketch the curve represented by the parametric equations. (b) Find a rectangular-coordinate equation for the curve by eliminating the parameter. $$x=\sec t, \quad y=\tan ^{2} t, \quad 0 \leq t<\pi / 2$$

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. $$x^{2}+4 y^{2}+20 x-40 y+300=0$$

(a) Use the discriminant to identify the conic. (b) Confirm your answer by graphing the conic using a graphing device. $$x^{2}-2 x y+3 y^{2}=8$$
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