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Find the standard equation of each parabola from the given information. Assume that the vertex is at the origin. Directrix is \(x=3\)

Short Answer

Expert verified
The equation is \(y^2 = -12x\).

Step by step solution

01

Understand the Parabola Orientation

In this problem, we have a parabola with the directrix given as a vertical line, \(x = 3\). This means the parabola opens horizontally, either to the left or right. Since the vertex is at the origin \((0,0)\) and the directrix is positive, the parabola will open to the left.
02

Identify the Distance to the Directrix

The distance \(p\) from the vertex \((0,0)\) to the directrix \(x = 3\) is 3 units. This represents the value of \(p\), which is a key component in writing the parabola's equation. Since the parabola opens to the left, \(p = -3\).
03

Write the Standard Equation of the Parabola

For a parabola that opens horizontally with vertex at the origin, the standard form is \(y^2 = 4px\). Substituting \(p = -3\) into the equation, we have:\[y^2 = 4(-3)x\]Simplifying this gives:\[y^2 = -12x\]This is the standard form equation of the given parabola.

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Key Concepts

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

Standard Equation of Parabola
When dealing with parabolas, understanding the standard equation is crucial. For a parabola with its vertex at the origin, the standard form can vary, depending on its orientation. A standard equation offers the simplest mathematical description of a parabola.

For a vertically oriented parabola (opening upward or downward), the standard form is \(x^2 = 4py\). Here, \(p\) represents the distance from the vertex to the focus or directrix. On the other hand, a horizontally oriented parabola (opening to the right or left) uses the equation \(y^2 = 4px\).

In our specific problem, because the parabola is horizontal and opens left due to the directrix at \(x = 3\), we use \(y^2 = 4px\) as the starting point. Understanding which equation form to use helps in easily determining the equation from known values of \(p\).
Vertex of a Parabola
The vertex is a fundamental characteristic of a parabola, acting as the point of symmetry. It's where the direction changes and is crucial for defining the curve's position.

In simple terms, the vertex is the turning point of the parabola. If you think of a parabola as a bowl, then the vertex would be the bottom of the bowl.

In this exercise, the vertex is given as the origin, \((0,0)\), which simplifies many calculations in understanding its geometry. Knowing the vertex's position allows us to explore other parts of the equation knowing this fixed point.
Horizontal Parabola
A horizontal parabola opens sideways - either right or left. This orientation depends primarily on the position of the directrix.

An important tip is if the directrix has an equation like \(x = k\), then the parabola is horizontal. When the vertex is at the origin and the directrix is x some positive value, like in our case with \(x = 3\), it confirms the parabola opens left.

Horizontal parabolas use the \(y^2 = 4px\) formula as discussed previously, making it essential to correctly identify the parabola's orientation based on provided information.
Directrix of Parabola
The directrix of a parabola is a straight line that, together with the focus, defines the set of points making up the parabola. It's involved in the fixed-distance property that characterizes parabolas.

In a horizontal parabola, the directrix is a vertical line. In this problem, it's represented by the equation \(x = 3\). This directrix is pivotal because the distance from the vertex to this line determines the value of \(4p\) in the standard equation \(y^2 = 4px\).

If the parabola opens to the left, \(p\) is negative, and we calculate the exact parabola equation from these known distances, as done here with \(y^2 = -12x\). The relationship between directrix and parabola equation highlights the interplay between algebra and geometry in conic sections.

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Most popular questions from this chapter

. Consider the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1\). (a) Show that its perimeter is $$ P=4 a \int_{0}^{\pi / 2} \sqrt{1-e^{2} \cos ^{2} t} d t $$ where \(e\) is the eccentricity. (b) The integral in part (a) is called an elliptic integral. It has been studied at great length, and it is known that the integrand does not have an elementary antiderivative, so we must turn to approximate methods to evaluate \(P .\) Do so when \(a=1\) and \(e=\frac{1}{4}\) using the Parabolic Rule with \(n=4\). (Your answer should be near \(2 \pi\). Why?) AS (c) Repeat part (b) using \(n=20\).

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