Finding The Directrix Of A Parabola Comprehensive Guide

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In the realm of conic sections, parabolas hold a special place, characterized by their unique properties and symmetrical U-shaped curves. Understanding the elements of a parabola, such as the focus, directrix, vertex, and axis of symmetry, is crucial for solving various mathematical problems and grasping their real-world applications. This article delves into the process of finding the equation of the directrix of a parabola, particularly when the equation is given in a general quadratic form. We will explore the underlying concepts, provide a step-by-step solution to a specific problem, and discuss the broader implications of parabolas in mathematics and beyond.

Parabolas are defined as the set of all points in a plane that are equidistant from a fixed point, called the focus, and a fixed line, called the directrix. The line passing through the focus and perpendicular to the directrix is the axis of symmetry, and the point where the parabola intersects the axis of symmetry is the vertex. The distance between the vertex and the focus is the same as the distance between the vertex and the directrix.

The general equation of a parabola can be expressed in several forms, depending on its orientation and position in the coordinate plane. A parabola with a vertical axis of symmetry has the equation

(x - h)^2 = 4p(y - k)

where (h, k) is the vertex and p is the distance between the vertex and the focus (and also the distance between the vertex and the directrix). Similarly, a parabola with a horizontal axis of symmetry has the equation

(y - k)^2 = 4p(x - h).

However, parabolas can also be represented by a more general quadratic equation of the form

Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0,

where not both A and C are zero. When B is non-zero, the parabola is rotated, making it more challenging to determine its key features directly from the equation. Our focus in this article is on finding the directrix of a parabola given an equation in this general form.

Let's consider the equation of a parabola given as:

9x^2 + 12xy + 4y^2 + 98x + 74y + 300 = 0.

The task is to find the equation of the directrix of this parabola. To solve this problem, we will go through a series of steps involving recognizing the parabolic form, determining the axis of the parabola, and then finding the directrix. The options provided are:

  1. 3x + 2y + 5 = 0

  2. 3x + 2y - 5 = 0

  3. 2x - 3y + 5 = 0

  4. 2x - 3y - 5 = 0

1. Recognizing the Parabolic Form

The given equation is in the general quadratic form:

Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0

where A = 9, B = 12, C = 4, D = 98, E = 74, and F = 300. To confirm that this equation represents a parabola, we can check the discriminant, which is given by:

B^2 - 4AC

For a parabola, the discriminant must be equal to zero. Plugging in the values, we get:

(12)^2 - 4(9)(4) = 144 - 144 = 0

Since the discriminant is zero, the equation indeed represents a parabola.

2. Identifying the Axis of the Parabola

Notice that the terms 9x^2, 12xy, and 4y^2 can be written as a perfect square:

9x^2 + 12xy + 4y^2 = (3x + 2y)^2

This suggests that the axis of the parabola is parallel to the line 3x + 2y = 0. The equation 3x + 2y = k represents a family of parallel lines, and the axis of the parabola is one of these lines.

3. Rewriting the Equation

Let's rewrite the given equation by completing the square. We can express the equation as:

(3x + 2y)^2 + 98x + 74y + 300 = 0

To proceed, we need to express the linear terms 98x + 74y in terms of 3x + 2y. Let u = 3x + 2y. Then, we can express the original equation in terms of u and another variable v, where v is perpendicular to u. A line perpendicular to 3x + 2y = 0 has the form 2x - 3y = 0. So, let v = 2x - 3y.

We now have two equations:

u = 3x + 2y
v = 2x - 3y

We can solve this system of equations for x and y in terms of u and v.

Multiply the first equation by 3 and the second equation by 2:

3u = 9x + 6y
2v = 4x - 6y

Add the two equations:

3u + 2v = 13x
x = (3u + 2v) / 13

Multiply the first equation by 2 and the second equation by 3:

2u = 6x + 4y
3v = 6x - 9y

Subtract the second equation from the first:

2u - 3v = 13y
y = (2u - 3v) / 13

Now, substitute x and y in the linear terms 98x + 74y:

98x + 74y = 98((3u + 2v) / 13) + 74((2u - 3v) / 13)

= (294u + 196v + 148u - 222v) / 13

= (442u - 26v) / 13

= 34u - 2v

So, the original equation becomes:

u^2 + 34u - 2v + 300 = 0

4. Completing the Square

Now, complete the square for the u terms:

(u^2 + 34u) - 2v + 300 = 0

(u^2 + 34u + 289) - 289 - 2v + 300 = 0

(u + 17)^2 = 2v - 11

5. Identifying the Vertex and the Parameter 'p'

We can rewrite the equation as:

(u + 17)^2 = 2(v - 11/2)

This is in the form U^2 = 4pV, where U = u + 17, V = v - 11/2, and 4p = 2, so p = 1/2. The vertex in the UV-plane is (U, V) = (-17, 11/2). The focus is at (U, V) = (-17, 11/2 + p) = (-17, 6), and the directrix is V = 11/2 - p = 11/2 - 1/2 = 5.

6. Finding the Directrix in the xy-Plane

The directrix in the UV-plane is V = 5. Substitute back v = 2x - 3y:

2x - 3y = 5

So, the equation of the directrix is 2x - 3y - 5 = 0.

Therefore, the equation of the directrix of the parabola 9x^2 + 12xy + 4y^2 + 98x + 74y + 300 = 0 is 2x - 3y - 5 = 0. This corresponds to option 4.

This step-by-step solution demonstrates the process of finding the directrix of a parabola given its general quadratic equation. The key steps involve recognizing the parabolic form, rewriting the equation by completing the square, identifying the vertex and parameter p, and finally, finding the equation of the directrix in the original coordinate system. Understanding these steps is essential for solving similar problems and gaining a deeper understanding of conic sections.

Parabolas are more than just mathematical curves; they have significant applications in various fields of science and engineering. Understanding their properties and equations is crucial for solving practical problems.

Real-World Applications

  1. Optics: Parabolic mirrors and lenses are used in telescopes, satellite dishes, and solar cookers. The parabolic shape allows parallel rays of light (or other electromagnetic radiation) to be focused at a single point, or conversely, a point source of light at the focus can be projected as a parallel beam. This property is vital in many optical devices.

  2. Antennas: Parabolic antennas are used in radar and communication systems. They function similarly to parabolic mirrors, focusing incoming signals at a receiver or transmitting signals in a focused beam. The precision of the parabolic shape ensures efficient signal transmission and reception.

  3. Projectile Motion: The path of a projectile (like a ball thrown in the air) under the influence of gravity approximates a parabola. Understanding the parabolic trajectory helps in predicting the range and height of projectiles, which is crucial in sports, military applications, and more.

  4. Architecture: Parabolic arches are structurally strong and aesthetically pleasing. They distribute weight evenly, making them ideal for bridges and buildings. Famous examples include the Gateway Arch in St. Louis, which is a catenary arch (a close relative of the parabola).

  5. Automotive Engineering: Headlights often use parabolic reflectors to create a focused beam of light. This design ensures that the light is projected forward efficiently, improving visibility for drivers.

Mathematical Significance

Parabolas are also important in mathematics for several reasons:

  1. Conic Sections: Parabolas are one of the four types of conic sections (the others being circles, ellipses, and hyperbolas). They are formed by the intersection of a plane and a double cone. Studying parabolas provides a foundation for understanding other conic sections and their properties.

  2. Quadratic Functions: Parabolas are the graphical representation of quadratic functions. Understanding the properties of parabolas helps in analyzing and solving quadratic equations and inequalities.

  3. Calculus: Parabolas are often used as examples in calculus to illustrate concepts such as derivatives, integrals, and optimization. The tangent to a parabola, for instance, has interesting properties that can be explored using calculus.

Tips for Solving Parabola Problems

  1. Recognize the Form: Identify whether the equation is in standard form or general form. Standard forms are easier to work with, so try to convert general forms to standard forms.

  2. Complete the Square: This technique is essential for rewriting quadratic equations in a form that reveals the vertex, focus, and directrix.

  3. Use the Discriminant: The discriminant B^2 - 4AC helps in identifying the type of conic section. For parabolas, it should be zero.

  4. Sketch the Parabola: Drawing a rough sketch can help visualize the problem and avoid errors. Label the vertex, focus, and directrix on your sketch.

  5. Practice Problems: Solve a variety of problems to become comfortable with different scenarios and techniques.

For those interested in delving deeper into the topic, here are some advanced concepts related to parabolas:

  1. Parametric Equations: Parabolas can be represented using parametric equations, which can be useful for describing the motion of a point along the curve.

  2. Focal Chord: A focal chord is a line segment passing through the focus of a parabola with endpoints on the parabola. The length of the focal chord perpendicular to the axis of symmetry (the latus rectum) is 4p, where p is the distance from the vertex to the focus.

  3. Tangent and Normal: The tangent to a parabola at a point bisects the angle between the line joining the point to the focus and the line perpendicular to the directrix. The normal is the line perpendicular to the tangent at that point.

  4. Applications in Physics: In physics, parabolas are used to describe the motion of projectiles, the shape of suspension cables in bridges, and the paths of charged particles in electric and magnetic fields.

In conclusion, understanding the properties and equations of parabolas is essential for various applications in mathematics, science, and engineering. This article provided a detailed guide on finding the equation of the directrix of a parabola, along with a comprehensive overview of the importance and applications of parabolas. By mastering these concepts, you can solve complex problems and appreciate the beauty and utility of this fundamental curve.

In this article, we addressed the problem of finding the directrix of a parabola given its general equation:

9x^2 + 12xy + 4y^2 + 98x + 74y + 300 = 0

We followed a step-by-step approach:

  1. Confirmed the equation represents a parabola using the discriminant.

  2. Identified the axis of the parabola by recognizing the perfect square form.

  3. Rewrote the equation using a transformation to a new coordinate system.

  4. Completed the square to bring the equation into standard form.

  5. Determined the vertex and the parameter p.

  6. Found the equation of the directrix in the new coordinate system and transformed it back to the original xy-plane.

The final equation of the directrix was found to be:

2x - 3y - 5 = 0

This comprehensive solution not only answers the specific question but also provides a detailed methodology for solving similar problems involving parabolas. The understanding of parabolas and their properties is crucial in various fields, including optics, antenna design, projectile motion analysis, and architecture. By mastering these concepts, one can tackle a wide array of mathematical and real-world challenges.