Problem 33 - Precalculus: Functions and Graphs

Question

Find a polar equation of the parabola with focus at the pole and the given vertex. $$V\left(4, \frac{\pi}{2}\right)$$

Step-by-Step Solution

Answer

The polar equation is \(r = \frac{4}{1 + \sin\theta}\).

Step 1: Understanding the Problem

To solve the problem, we need to find the polar equation of a parabola with its focus at the pole \((0, 0)\) and its vertex given in polar coordinates \((4, \frac{\pi}{2})\). Polar equations for conics can be expressed in the form \(r = \frac{ed}{1 - e\cos\theta}\) or \(r = \frac{ed}{1 + e\cos\theta}\), where \(e\) is the eccentricity and \(d\) is the directrix.

Step 2: Identify Eccentricity and Directrix Distance

For a parabola, the eccentricity \(e = 1\). The given vertex \(V(4, \frac{\pi}{2})\) is at a distance of \(4\) from the pole vertically upwards. This means the vertex is above the focus; thus, the parabola opens upwards. The directrix, therefore, is horizontal.

Step 3: Use Polar Equation for Vertical Parabola

For a parabola that opens upwards or downwards, the polar equation can be written as \(r = \frac{d}{1 + e\sin\theta}\) or \(r = \frac{d}{1 - e\sin\theta}\). Here, since the vertex is at \(\theta = \frac{\pi}{2}\), we use \(r = \frac{d}{1 + \sin\theta}\).

Step 4: Calculate Directrix Distance

Since the vertex \((4, \frac{\pi}{2})\) is at a distance of \(4\) from the focus and lies along the directrix horizontally, \(d = 4\).

Step 5: Substitute Values into the Equation

Substitute \(d = 4\) and \(e = 1\) into the polar equation of the conic. The equation becomes \(r = \frac{4}{1 + \sin\theta}\).

Key Concepts

Parabola Focus at the Pole

In polar coordinates, when we say that a parabola has its focus at the pole, we mean that the focus of the parabola is located at the origin point in the polar coordinate system. The focus is a special point used to define the parabola, alongside the directrix, its guiding line.
For parabolas, the eccentricity is fixed at 1. This eccentricity tells us that all points on the parabola are equidistant from the focus and the directrix. When the focus is at the pole, we adjust the polar equations to account for this central positioning. Specifically, the focus's location at (0,0) allows us to use a simplified form when determining the polar equation since there's symmetry about the central point. Always keep in mind that the parabola will symmetrically face away from the directrix when the focus is directly at the pole.

Polar Coordinates

Polar coordinates offer a way of representing points in a plane using a distance and an angle. Unlike Cartesian coordinates, which use horizontal and vertical distances from an origin point, polar coordinates describe a location with the length of the radius vector (distance from the pole) and the angle from a reference direction (usually the positive x-axis).

• Poles: The center point, also known as the origin in Cartesian coordinates.
• Radius (r): The straight line distance from the pole to a point.
• Angle (θ): The angle measured from the reference direction to the radius vector.

In the problem, knowing that the vertex is at (4, \(\frac{\pi}{2}\)) tells us that the vertex is 4 units away from the pole at an angle of \(90^{\circ}\) or \(\frac{\pi}{2}\) radians, revealing the upward orientation of the parabola.

Eccentricity of a Conic

Eccentricity is a crucial concept when dealing with conics like circles, ellipses, parabolas, and hyperbolas.

• Circles: Have eccentricity \(e = 0\).
• Ellipses: Have eccentricity \(0 < e < 1\).
• Parabolas: Always have eccentricity \(e = 1\).
• Hyperbolas: Have eccentricity \(e > 1\).

For parabolas, being a special type of conic section, the eccentricity's value of 1 means all points on a parabola are equidistant from the focus and the directrix. This unique characteristic helps us easily identify the nature of the curve as a parabola and simplify calculationsdata depending on the symmetry and simplicity of having the focus at the pole and using polar coordinates.

Directrix in Polar Equations

The directrix is a line used to help define conic sections. For a parabola, it is positioned such that each point on the parabola is equidistant to the focus and the directrix.
When developing a polar equation for a parabola, the directrix plays a crucial role, especially with the focus at the pole. Its orientation affects the form of the polar equation:

• Horizontal directrix for a vertical facing parabola often uses \(r = \frac{d}{1 \pm \sin\theta}\).
• Vertical directrix for a horizontal facing parabola uses \(r = \frac{d}{1 \pm \cos\theta}\).

In the given problem, since the vertex is at \((4, \frac{\pi}{2})\), the directrix is positioned horizontally, and we use the form \(r = \frac{d}{1 + \sin\theta}\), where \(d\) is the distance to the directrix.