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Chapter 7: Problem 71

Give the focus, directrix, and axis of each parabola. $$y^{2}=\frac{1}{16} x$$

Short Answer

Expert verified
Focus: \((\frac{1}{64}, 0)\), Directrix: \(x = -\frac{1}{64}\), Axis: \(y = 0\).

Step by step solution

01

Identify the General Form of the Parabola

The given parabola equation is \( y^2 = \frac{1}{16}x \). This is in the form \( y^2 = 4px \), where \( p \) is the distance from the vertex to the focus or directrix.
02

Solve for p

Compare \( y^2 = \frac{1}{16}x \) to the standard form \( y^2 = 4px \). Therefore, \( 4p = \frac{1}{16} \), so \( p = \frac{1}{16} \div 4 = \frac{1}{64} \). Thus, \( p = \frac{1}{64} \).
03

Determine the Vertex

Since the equation is in the form \( y^2 = 4px \), the vertex of this parabola is at the origin, \((0, 0)\).
04

Find the Focus

In the form \( y^2 = 4px \), the focus is to the right of the vertex at \((p, 0)\), which is \((\frac{1}{64}, 0)\).
05

Find the Directrix

The directrix of the parabola \( y^2 = 4px \) is a vertical line, and it will be \( x = -p \). Therefore, the directrix is \( x = -\frac{1}{64} \).
06

Determine the Axis

Since the parabola opens to the right, the axis of symmetry is horizontal at \( y = 0 \).

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Key Concepts

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

Focus of Parabola
The focus of a parabola is a key point that helps in defining the shape and orientation of the parabola. For a parabola in the form \(y^2 = 4px\), the focus is found by moving from the vertex a distance \(p\) in the direction the parabola opens.
In our example, the parabola is described by the equation \(y^2 = \frac{1}{16}x\), which matches the standard form with \(4p = \frac{1}{16}\). Solving for \(p\), we find \(p = \frac{1}{64}\).
The parabola opens to the right since the\(x\) term is positive, placing the focus at \(\left(\frac{1}{64}, 0\right)\), precisely \(\frac{1}{64}\) units to the right of the vertex, which is at the origin \((0, 0)\).
  • The focus is crucial as it directs all light rays reflecting off the parabola's surface to converge at this point.
Directrix of Parabola
The directrix of a parabola is a line that helps define the set of points making up the parabola. The parabola is the locus of points such that each point on the curve is equidistant from the focus and the directrix. For the given equation, \(y^2 = \frac{1}{16}x\), which fits the form \(y^2 = 4px\), we have \(p = \frac{1}{64}\).
In this orientation, the directrix is a vertical line, positioned opposite to the focus, and located \(p\) units from the vertex along the \(-x\)-axis direction. Hence, the directrix is at \(x = -\frac{1}{64}\).
  • The directrix provides a reference line used, along with the focus, to define the curve geometrically.
Axis of Parabola
The axis of a parabola is the line that runs symmetrically through the vertex and focus. It is essentially the "backbone" of the parabola along which it is symmetric. For our parabola, which is in the standard form \(y^2 = 4px\), the axis is horizontal, parallel to the \(x\)-axis.
Since the parabola opens to the right, its axis is the line \(y = 0\). This line passes through the vertex at \((0, 0)\), which is the center of symmetry for the parabola.
  • The axis helps in visualizing and understanding the symmetry of the parabola, as all points are mirrored across this line.

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Most popular questions from this chapter

Find an equation for each hyperbola. \(y\) -intercepts \((0, \pm 5) ;\) foci \((0, \pm 3 \sqrt{3})\)

Graph each parabola by hand, and check using a graphing calculator. Give the vertex, axis, domain, and range. $$x=\frac{2}{3} y^{2}-4 y+8$$

Path of an Object on a Planet When an object moves under the influence of a gravitational force (without air resistance), its path can be parabolic. This is the path of a ball thrown near the surface of a planet or other celestial object. Suppose two balls are simultaneously thrown upward at a \(45^{\circ}\) angle on two different planets. If their initial velocities are both \(30 \mathrm{mph}\), then their \(x y\) -coordinates in feet can be expressed by the equation $$ y=x-\frac{g}{1922} x^{2} $$ where \(g\) is the acceleration due to gravity. The value of \(g\) will vary with the mass and size of the planet. (Source: Zeilik, M., S. Gregory, and E. Smith, Introductory Astronomy and Astrophysics, Saunders College Publishers.) (a) On Earth, \(g=32.2\) and on Mars, \(g=12.6 .\) Find the two equations, and use the same screen of a graphing calculator to graph the paths of the two balls thrown on Earth and Mars. Use the window [0,180] by \([0,120] .\) (Hint: If possible, set the mode on your graphing calculator to simultaneous.) (b) Determine the difference in the horizontal distances traveled by the two balls.

Find an equation for each hyperbola. Foci \((-3 \sqrt{5}, 0)\) and \((3 \sqrt{5}, 0)\); asymptotes \(y=\pm 2 x\)

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