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Chapter 11: Problem 28

Exer. 19-30: Find an equation of the parabola that satisfies the given conditions. Vertex at the origin, symmetric to the \(y\)-axis, and passing through the point \((6,3)\)

Short Answer

Expert verified
The equation of the parabola is \( y = \frac{1}{12}x^2 \).

Step by step solution

01

Understanding the Parabola

Since the parabola is symmetric to the y-axis, its equation will be of the form \( y = ax^2 \). Given that the vertex is at the origin \((0,0)\), the vertex form of the equation is simplified to this standard form.
02

Substitute the Point

The parabola passes through the point \((6,3)\). Substitute \(x = 6\) and \(y = 3\) into the equation \( y = ax^2 \): \[ 3 = a(6)^2 \]
03

Solve for 'a'

Solve the equation \( 3 = a(6)^2 \) to find the value of \(a\): \[ 3 = 36a \] Divide both sides by 36: \[ a = \frac{3}{36} = \frac{1}{12} \]
04

Write the Equation of the Parabola

Now that we know \(a = \frac{1}{12}\), the equation of the parabola is: \[ y = \frac{1}{12}x^2 \]

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Key Concepts

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

Vertex Form
Parabolas have different forms of equations, and one of them is the vertex form. When the vertex of the parabola is at the origin, we simplify the vertex form. The general vertex form equation is: \[ y = a(x-h)^2 + k \]Here,
  • \( (h, k) \) is the vertex of the parabola.
  • \( a \) is a constant that affects the opening and orientation of the parabola.
For a vertex at the origin \((0, 0)\), the equation simplifies to:\[ y = ax^2 \]This form is straightforward as there are no extra terms; the vertex is simply the point where the parabola touches the line axis(both x and y are zero here).
Therefore, when you see a parabola's vertex at the origin in a problem, it makes the equation much simpler to handle.
Symmetry
Symmetry is a key feature in parabolas, making them easier to understand and work with. A parabola symmetrical to the \(y\)-axis looks the same on both sides of this axis. Imagine drawing a vertical line right down the middle; the two halves mirror each other.
This symmetry means our equation will be an "even" function, such as \(y = ax^2\), as opposed to an equation like \(y = ax^2 + bx + c\), which can describe a parabola offset from the symmetry.
  • When the symmetry is around the y-axis, the function omits the linear \(bx\) term.
  • Instead, \(y = ax^2\) shows each \(x\) value has corresponding \(-x\) outputting the same \(y\).
Therefore, the equation of our parabola maintains this symmetrical property through its simplistic form.
Substitute Point
Substituting a point into the parabola's equation is a method for identifying unknown constants. If our parabola passes through a point, all we need to do is plug that point's coordinates into the equation. In this problem, the point \((6, 3)\) provides crucial information to find "a."
  • Start with the equation: \(y = ax^2\)
  • Substitute \(x = 6\) and \(y = 3\)
Doing this gives us:\[ 3 = a(6)^2 \]Thus, substituting known values from a point directly into your equation helps to adjust or find the specific parameters of the parabola.
Solve for a
Solving for the constant "a" is a fundamental step when finding the specific equation of a parabola based on given conditions. Once substitution of a point is complete, you have an equation you can solve.
Following the previous step where \(3 = a \times 36\):
  • Rearrange to isolate \(a\).
  • Divide both sides by 36.
This yields:\[ a = \frac{3}{36} = \frac{1}{12} \]Now, the value \(a = \frac{1}{12}\) describes how wide or narrow the parabola will be. It indicates that for every 1 unit increase in \(x\), the \(y\) value scales by the factor of \( \frac{1}{12} x^2 \). Therefore, once "a" is known, it gets plugged back into the equation, finalizing the expression for the parabola like so:\[ y = \frac{1}{12}x^2 \]

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Most popular questions from this chapter

A radio telescope has the shape of a paraboloid of revolution, with focal length \(p\) and diameter of base \(2 a\). From calculus, the surface area \(S\) available for collecting radio waves is $$ S=\frac{8 \pi p^{2}}{3}\left[\left(1+\frac{a^{2}}{4 p^{2}}\right)^{3 / 2}-1\right] $$ One of the largest radio telescopes, located in Jodrell Bank, Cheshire, England, has diameter 250 feet and focal length 75 feet. Approximate \(S\) to the nearest thousand square feet.

A satellite antenna dish has the shape of a paraboloid that is 10 feet across at the open end and is 3 feet deep. At what distance from the center of the dish should the receiver be placed to receive the greatest intensity of sound waves?

Exer. 1-12: Find the eccentricity, and classify the conic. Sketch the graph, and label the vertices. $$ r=\frac{3}{2+2 \cos \theta} $$

A satellite will travel in a parabolic path near a planet if its velocity \(v\) in meters per second satisfies the equation \(v=\sqrt{2 k / r}\), where \(r\) is the distance in meters between the satellite and the center of the planet and \(k\) is a positive constant. The planet will be located at the focus of the parabola, and the satellite will pass by the planet once. Suppose a satellite is designed to follow a parabolic path and travel within 58,000 miles of Mars, as shown in the figure. (a) Determine an equation of the form \(x=a y^{2}\) that describes its flight path. (b) For Mars, \(k=4.28 \times 10^{13}\). Approximate the maximum velocity of the satellite. (c) Find the velocity of the satellite when its \(y\)-coordinate is 100,000 miles.

Exer. 25-32: Find a polar equation of the conic with focus at the pole that has the given eccentricity and equation of directrix. $$ e=\frac{4}{3}, \quad r \cos \theta=-3 $$
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