Problem 66 - Thomas Calculus

Question

In Exercises \(65-68\) , find a polar equation for the given curve. In each case, sketch a typical curve. $$ y^{2}=4 a x+4 a^{2} $$

Step-by-Step Solution

Answer

Convert to polar: \( r = \frac{4a \cos \theta \pm \sqrt{16a^2 \cos^2 \theta + 4a^2 \sin^2 \theta}}{2 \sin^2 \theta} \).

Step 1: Identify the given equation type

The given equation \( y^2 = 4ax + 4a^2 \) is a conic section. It can be recognized as a parabola because it is in the form \( y^2 = 4px \), where \( p \) is a parameter.

Step 2: Rearrange the equation

Rearrange the equation \( y^2 = 4ax + 4a^2 \) to the form \( y^2 = 4ax + 4a^2 \) which becomes \( y^2 = 4a(x + a) \). This gives a clearer representation of the parabola's structure.

Step 3: Convert Cartesian to polar coordinates

In polar coordinates, \( x = r \cos \theta \) and \( y = r \sin \theta \). Substitute these into the rewritten equation: \( (r \sin \theta)^2 = 4a(r \cos \theta + a) \).

Step 4: Simplify the polar equation

The substitution yields \( r^2 \sin^2 \theta = 4ar \cos \theta + 4a^2 \). Rearrange to isolate \( r \): \( r^2 \sin^2 \theta - 4ar \cos \theta - 4a^2 = 0 \). This is a quadratic in \( r \).

Step 5: Solve the quadratic equation for \( r \)

Use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = \sin^2 \theta \), \( b = -4a \cos \theta \), and \( c = -4a^2 \). Solve for \( r \) to get the polar form of the parabola.

Step 6: Sketch the curve

With the polar equation, sketch the graph of the parabola in polar coordinates, confirming that it opens to the right as per the original Cartesian form.

Key Concepts

Conic Sections

Conic sections are curves obtained by intersecting a plane with a cone. They form some of the most basic geometric shapes:

• Circle
• Ellipse
• Parabola
• Hyperbola

Understanding the nature of conic sections is important because they appear in many fields such as astronomy, engineering, and physics.
Each section has specific characteristics that define its shape and behavior. For instance, a circle is a special case of the ellipse where the two foci coincide.
A parabola, on the other hand, is formed when the plane is parallel to the cone's side. It focuses more on direction rather than a central point, unlike ellipses and circles. Hyperbolas are characterized by their two branches that open in opposite directions.

Parabola

A parabola is a distinct type of conic section that can be defined as the set of all points equidistant from a point (called the focus) and a line (called the directrix).
In Cartesian coordinates, the standard form of a parabola opening to the right is given by \(y^2 = 4px\). Here, 'p' represents the distance from the vertex to the focus, which determines how wide or narrow the parabola appears.
Parabolas are widely seen in real life scenarios. Examples include the path of a thrown ball (projectile motion) or the design of satellite dishes and car headlights, which utilize their reflective properties.

Coordinate Conversion

Coordinate conversion involves transforming one coordinate system to another, such as converting Cartesian coordinates to polar coordinates. This is essential for analyzing shapes that are naturally described in a radial or circular form.
The link between the two systems is established by the equations:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)
• \(r = \sqrt{x^2 + y^2}\)
• \(\theta = \tan^{-1}(\frac{y}{x})\)

By substituting these expressions into an equation in one form, you can find its equivalent in the other form. This tool is particularly useful for equations of curves that are easier to analyze or sketch in one coordinate system versus another.

Polar Equation

A polar equation describes a curve using polar coordinates, where each point on the curve is determined by a radius and an angle with respect to the origin.
For the parabola in the exercise, converting its Cartesian form \(y^2 = 4ax + 4a^2\) into a polar form involves applying the relationships between Cartesian and polar systems. The conversion results in a polar equation that can reveal new insights into the curve's shape and behavior in a radial context.
Upon deriving the polar equation, the task is to solve for \(r\). The quadratic nature of the resulting equation requires careful manipulation, often using the quadratic formula, to isolate \(r\).
This results in a concise polar expression that can be sketched or analyzed further, providing a deeper understanding of the parabola's geometric properties in a polar framework.