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Chapter 11: Problem 33

Find the equation of the parabola satisfying the given conditions. In each case, assume that the vertex is at the origin. The parabola is symmetric about the \(x\) -axis, the \(x\) -coordinate of the focus is negative, and the length of the focal chord perpendicular to the \(x\) -axis is 9

Short Answer

Expert verified
The equation of the parabola is \( y^2 = -9x \).

Step by step solution

01

Identify the Parabola Type

Since the parabola is symmetric about the x-axis, it opens either to the left or the right. Furthermore, the problem states that the x-coordinate of the focus is negative, indicating the parabola opens to the left.
02

Determine the Parabola's Equation Form

A parabola that opens to the left with its vertex at the origin has the form \( y^2 = -4px \), where \( p > 0 \) is the distance from the vertex to the focus along the x-axis.
03

Use Focal Chord Condition

The focal chord perpendicular to the x-axis (a vertical line segment through the focus) has length 9. The length of a focal chord perpendicular to the axis of symmetry of a parabola \( y^2 = -4px \) is given by \( 4p \). Set \( 4p = 9 \).
04

Solve for p

From the equation \( 4p = 9 \), solve for \( p \) to get \( p = \frac{9}{4} \).
05

Write the Final Equation

Substitute \( p = \frac{9}{4} \) into the parabola's equation \( y^2 = -4px \) to get \( y^2 = -4\left(\frac{9}{4}\right)x = -9x \).

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Key Concepts

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

Vertex Form
When discussing parabolas, the vertex form is essential for describing the shape and position. In general, the vertex form of a parabola's equation is expressed as
  • For a vertical parabola: \( y = a(x-h)^2 + k \)
  • For a horizontal parabola: \( x = a(y-k)^2 + h \)
Here,
  • \( (h, k) \) represents the vertex of the parabola, which is a crucial point where the parabola changes direction.
  • For vertical parabolas, the axis of symmetry is the line \( x = h \).
  • For horizontal parabolas, the axis of symmetry is the line \( y = k \).
In our exercise, since the vertex is at the origin
  • The equation simplifies dramatically to either \( y^2 = 4px \) or \( x^2 = 4py \) depending on whether the parabola opens horizontally or vertically, respectively.
Understanding where the vertex is located helps in predicting the parabola's symmetry and direction.
Focus and Directrix
The focus and directrix are pivotal in defining a parabola. They provide a way to understand how a parabola behaves.
  • The focus is a point from which distances are measured to determine the distance of any point on the parabola.
  • The directrix is a line that serves as a reference point for these distances.
Mathematically, any point \( (x, y) \) on the parabola is equidistant from the focus \( (h \, \pm \, p, k) \) and the directrix, which is the line \( x = h \pm p \) or \( y = k \mp p \) for horizontal or vertical parabolas respectively.
In the exercise, with the parabola opening to the left,
  • The focus is located at \( igg(-\frac{9}{4}, 0\bigg) \).
  • The corresponding directrix is a vertical line \( x = \frac{9}{4} \).
Focal Chord
A focal chord is a line segment that passes through the focus of a parabola and has its endpoints on the parabola.
For lines perpendicular to the axis of symmetry, these are particularly interesting due to properties they hold.
  • This property leveraged in problem solving, especially in our exercise, provided the necessary condition to determine \( p \).
The parabola equation \( y^2 = -4px \) has its axis of symmetry along the \( x \)-axis with the vertex at the origin.
  • For such parabolas, the length of a focal chord perpendicular to the axis of symmetry is simply \( |4p| \).
  • Incorporating our given condition \( 4p = 9 \), we tweaked this relationship to find the value of \( p \).
This connection between the focal chord and \( p \) helps in converting physical descriptions into an equation.
Symmetry of Parabolas
Parabolas are well-known for their symmetric properties. A parabola's symmetry can tell us much about its properties.
  • The axis of symmetry is a line that divides the parabola into two mirror-image halves.
  • For parabolas aligned to the x-axis, the symmetry line is horizontal.
  • For vertical parabolas, it's vertical over the y-axis.
In our case, with a parabola that opens to the left:
The symmetry is about the \( x \)-axis, ensuring that for each point \( (x, y) \) on the parabola, there is a corresponding point \( (x, -y) \). This symmetry helps simplify calculations and visualizations of the parabola's trajectory and focus position vividly.

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Most popular questions from this chapter

In this exercise we consider how the eccentricity \(e\) influences the graph of an ellipse \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1.\) (a) For simplicity, we suppose that \(a=1\) so that the equation of the ellipse is \(x^{2}+\left(y^{2} / b^{2}\right)=1 .\) Solve this equation for \(y\) to obtain $$y=\pm b \sqrt{1-x^{2}}$$ (b) Assuming that \(a=1,\) show that \(b\) and \(e\) are related by the equation \(b^{2}=1-e^{2},\) from which it follows that \(b=\pm \sqrt{1-e^{2}} .\) The positive root is appropriate here because \(b>0 .\) Thus, we have $$b=\sqrt{1-e^{2}}$$ (c) Using equation (2) to substitute for \(b\) in equation (1) yields \(y=\sqrt{1-e^{2}} \sqrt{1-x^{2}} \quad\) or \(\quad y=-\sqrt{1-e^{2}} \sqrt{1-x^{2}}\) This pair of equations represents an ellipse with semimajor axis 1 and eccentricity \(e .\) Using the value \(e=0.3,\) graph equations ( 3 ) in the viewing rectangle [-1,1] by [-1,1] Use true proportions and, for comparison, add to your picture the circle with radius 1 and center \((0,0) .\) Note that the ellipse is nearly circular. (d) Follow part (c) using \(e=0.017 .\) This is approximately the eccentricity for Earth's orbit around the Sun. How does the ellipse compare to the circle in this case? (e) Follow part (c) using, in turn, \(e=0.4, e=0.6, e=0.8\) \(e=0.9, e=0.99,\) and \(e=0.999 .\) Then, in complete sentences, summarize what you've observed.

(a) Verify that the points \(A(5,1), B(4,-2),\) and \(C(-1,3)\) all lie on the ellipse \(x^{2}+3 y^{2}=28\). (b) Find a point \(D\) on the ellipse such that \(\overline{C D}\) is parallel to \(\overline{A B}\). (c) If \(O\) denotes the center of the ellipse, show that the triangles \(O A C\) and \(O B D\) have equal areas. Suggestion: In computing the areas, the formula given at the end of Exercise 34 in Section 1.4 is useful.

Find \(\sin \theta\) and \(\cos \theta,\) where \(\theta\) is the (acute) angle of rotation that eliminates the \(x^{\prime} y^{\prime}\) -term. Note: You are not asked to graph the equation. $$161 x y-240 y^{2}-1=0$$

Graph each conic section. If the conic is a parabola, specify (using rectangular coordinates) the vertex and the directrix. If the conic is an ellipse, specify the center, the eccentricity, and the lengths of the major and minor axes. If the conic is a hyperbola, specify the center, the eccentricity, and the lengths of the transverse and conjugate axes. $$r=\frac{2}{\sqrt{2}-\sqrt{2} \sin \theta}$$

Graph the equations. $$x^{2}+y^{2}=2 x y+4 x+4 y-8$$
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