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Chapter 11: Problem 13

Use the given information to find the equation of the hyperbola. Write the answer in the form \(A x^{2}-B y^{2}=C.\) The foci are \((\pm 2,0),\) and the directrices are \(x=\pm 1.\)

Short Answer

Expert verified
The equation of the hyperbola is \(x^2 - y^2 = 2\).

Step by step solution

01

Identify Key Elements

Given that the foci of the hyperbola are \((\pm 2, 0)\), we can immediately infer that the hyperbola is oriented horizontally. Hence, the center is at the origin \((0,0)\). The general equation for such a hyperbola is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).
02

Understand the Relationship Between Foci and Directrices

For a hyperbola, the distance from its center to one of its foci is represented by \(c\), and the directrix distance \(a^2/c\). Given foci \((\pm 2,0)\), we find \(c = 2\). The directrix equation \(x = \pm 1\) implies \(a^2/c = 1\).
03

Calculate \(a\) and \(b\)

Using the directrix relationship, \(a^2 = c = 2\), so \(a = \sqrt{2}\). To find \(b\), use \(c^2 = a^2 + b^2\). Thus, \(4 = 2 + b^2\). Solving for \(b^2\) gives \(b^2 = 2\).
04

Write the Equation of the Hyperbola

We now have all the necessary values: \(a^2 = 2\) and \(b^2 = 2\). Substitute these into the standard form equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), which simplifies to \(\frac{x^2}{2} - \frac{y^2}{2} = 1\). Multiply both sides by 2 to get the desired form: \(x^2 - y^2 = 2\).

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Key Concepts

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

Foci of Hyperbola
The foci of a hyperbola are two distinct points that play a crucial role in its shape and orientation. The hyperbola is defined by the distances from any point on the hyperbola to each of these foci. For the given problem, the foci are located at \((\pm 2, 0)\).
  • This arrangement means that the center of the hyperbola is at the origin, \((0,0)\).
  • In this situation, the foci lie along the x-axis, a sure sign that the hyperbola is horizontally oriented.
  • The distance from the center to each focus is denoted by \(c\), which in our problem is equal to 2, derived from the coordinates \((2,0)\).
Understanding the placement of the foci helps determine other important features of the hyperbola, such as its orientation and the location of its directrices.
Directrices of Hyperbola
The directrices of a hyperbola are lines that are used along with the foci to rigorously define the curve. For every point on the hyperbola, the ratio of its distance to the focus and its distance to the corresponding directrix remains consistent.
  • In this problem, the directrices are given as \(x = \pm 1\), indicating that they are also aligned with the orientation of the hyperbola.
  • The defining property of these lines, \(x = \pm 1\), allows one to calculate \(a\), the semi-major axis, using the equation \(a^2/c = 1\).
  • Since the foci are located at \((\pm 2,0)\), we know \(c = 2\).
  • By substituting \(c\) into the relationship \(a^2/c = 1\), we find \(a^2 = 2\), thus \(a = \sqrt{2}\).
These calculations lead to determining the parameters essential for expressing the hyperbola in its standard form equation.
Hyperbola Orientation
The orientation of a hyperbola can either be horizontal or vertical, depending entirely upon the positions of its foci. For the hyperbola given in the exercise, we determined that it is horizontally oriented due to its foci placement at \((\pm 2, 0)\).
  • A horizontal orientation means the transverse axis is aligned along the x-axis.
  • As implied by the exercise, the equation takes on the format: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).
  • Indentifying the orientation is essential before proceeding with further calculations and formulizing the equation of the hyperbola.
Understanding the orientation also sets the stage for treating other calculations, such as solving for 'a' and 'b', which result from the equations.
Hyperbola Standard Form
The standard form of a hyperbola allows for a clear and complete representation of the curve's properties. Knowing whether the hyperbola is horizontally or vertically oriented guides this formulation. For a horizontally oriented hyperbola, as is the case in this exercise, the standard form is written as:
  • \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
  • With \(a^2 = 2\) and \(b^2 = 2\), those values substitute into the standard form, making it \(\frac{x^2}{2} - \frac{y^2}{2} = 1\).
This equation is identical on both sides until multiplied by 2 to align with the required format of the original exercise problem, resulting in:
  • \(x^2 - y^2 = 2\)
This conversion involves an understanding of hyperbola fundamentals, particularly how to transition between different expression formats.

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Most popular questions from this chapter

This exercise outlines the steps required to show that the equation of the tangent to the ellipse \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1\) at the point \(\left(x_{1}, y_{1}\right)\) on the ellipse is \(\left(x_{1} x / a^{2}\right)+\left(y_{1} y / b^{2}\right)=1\) (a) Show that the equation \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1\) is equivalent to $$b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$$(1) Conclude that \(\left(x_{1}, y_{1}\right)\) lies on the ellipse if and only if $$b^{2} x_{1}^{2}+a^{2} y_{1}^{2}=a^{2} b^{2}$$(2) (b) Subtract equation ( \(2)\) from equation (1) to show that $$b^{2}\left(x^{2}-x_{1}^{2}\right)+a^{2}\left(y^{2}-y_{1}^{2}\right)=0$$(3) Equation ( 3 ) is equivalent to equation ( 1 ) provided only that \(\left(x_{1}, y_{1}\right)\) lies on the ellipse. In the following steps, we will find the algebra much simpler if we use equation ( 3 ) to represent the ellipse, rather than the equivalent and perhaps more familiar equation (1). (c) Let the equation of the line tangent to the ellipse at \(\left(x_{1}, y_{1}\right)\) be $$y-y_{1}=m\left(x-x_{1}\right)$$ Explain why the following system of equations must have exactly one solution, namely, \(\left(x_{1}, y_{1}\right);\) $$\left\\{\begin{array}{ll}b^{2}\left(x^{2}-x_{1}^{2}\right)+a^{2}\left(y^{2}-y_{1}^{2}\right)=0 & (4) \\ y-y_{1}=m\left(x-x_{1}\right) & (5)\end{array}\right.$$ (d) Solve equation ( 5 ) for \(y\), and then substitute for \(y\) in equation (4) to obtain \(b^{2}\left(x^{2}-x_{1}^{2}\right)+a^{2} m^{2}\left(x-x_{1}\right)^{2}+2 a^{2} m y_{1}\left(x-x_{1}\right)=0\) (e) Show that equation (6) can be written \(\left(x-x_{1}\right)\left[b^{2}\left(x+x_{1}\right)+a^{2} m^{2}\left(x-x_{1}\right)+2 a^{2} m y_{1}\right]=0\) (f) Equation (7) is a quadratic equation in \(x,\) but as was pointed out earlier, \(x=x_{1}\) must be the only solution. (That is, \(x=x_{1}\) is a double root.) Thus the factor in brackets must equal zero when \(x\) is replaced by \(x_{1}\) Use this observation to show that $$m=-\frac{b^{2} x_{1}}{a^{2} y_{1}}$$ This represents the slope of the line tangent to the ellipse at \(\left(x_{1}, y_{1}\right)\) (g) Using this value for \(m,\) show that equation (5) becomes $$b^{2} x_{1} x+a^{2} y_{1} y=b^{2} x_{1}^{2}+a^{2} y_{1}^{2}$$(8) (h) Now use equation (2) to show that equation (8) can be written $$\frac{x_{1} x}{a^{2}}+\frac{y_{1} y}{b^{2}}=1$$ which is what we set out to show.

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