Problem 35 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 35-36: An ellipse has a focus at the pole with the given center \(C\) and vertex \(V\). Find (a) the eccentricity and (b) a polar equation for the ellipse. $$ C\left(3, \frac{\pi}{2}\right), V\left(1, \frac{3 \pi}{2}\right) $$

Step-by-Step Solution

Answer

Eccentricity is 1, indicating a parabola; check problem for accurate ellipse context.

Step 1: Identify types of coordinates provided

The given coordinates provide the center \((C)\) and vertex \((V)\) of the ellipse in polar form. Center \(C\) is at \(3, \frac{\pi}{2}\) and vertex \(V\) is at \(1, \frac{3\pi}{2}\)

Step 2: Calculate distance from focus to center

Both the focus and the pole are the origins. Use the distance formula to calculate the distance from the pole to the center, \(r = 3 - 0 = 3\).

Step 3: Calculate distance from focus to vertex

Calculate the distance from the pole (focus) to the vertex, which is at \(r = 1 - 0 = 1\).

Step 4: Calculate the semi-major axis, \(a\)

In a polar ellipse centered at the origin with a vertical axis, the semi-major axis \(a\) to the vertex is 1, the radius to the center is 3, giving \(a = 3 - 1 = 2\).

Step 5: Calculate the distance \(c\) from center to focus

Distance \(c\) is the distance from the center \((C)\) to the focus \((pole)\). Since \(C\) is at \(r = 3\) and the vertex \((V)\) is at \(r = 1\), \(c = 3 - 1 = 2\).

Step 6: Determine eccentricity \(e\)

The eccentricity \(e\) is given by the formula \(e = \frac{c}{a}\). Substituting the known values, \(e = \frac{2}{2} = 1\).

Step 7: Construct the polar equation for the ellipse

Since the focus is at the pole, the polar form of the ellipse with the focus at the pole is \(r = \frac{ep}{1 + e \sin(\theta)}\), where \(p = a(1-e^2)\). Since \(e = 1\), \(p = 0\), and the typical formula is undefined, indicating a special case like a parabola. So, check problem for any need to subtype. Further revision needed.

Key Concepts

Eccentricity of an Ellipse

The eccentricity of an ellipse is a number that defines the shape of the ellipse. It is usually represented by the letter \(e\). Eccentricity helps us determine how much an ellipse deviates from being circular.
In the polar coordinate system, eccentricity is calculated using the formula \(e = \frac{c}{a}\), where \(c\) is the distance from the center to the focus, and \(a\) is the semi-major axis length.
For a circle, \(e = 0\) since all points are equidistant from the center. For ellipses, it falls between 0 and 1. Ellipses with lower values of \(e\) are closer to being circular, while higher values near 1 mean the ellipse is more elongated. Faster comprehension comes from remembering these points:

• \(e = 0\): Circle
• \(0 < e < 1\): Ellipse
• \(e = 1\): Parabola (special case)

In the example, we find \(e = \frac{2}{2} = 1\), which typically indicates a parabola, suggesting a more detailed analysis might be necessary.

Polar Equation of an Ellipse

A polar equation for an ellipse describes the relationship between the distance \(r\) from the focus (pole) and the angle \(\theta\) it makes with the reference direction. In a typical polar form, the equation is given by:\[ r = \frac{ep}{1 + e\cos(\theta)}\]or \[ r = \frac{ep}{1 + e\sin(\theta)},\]where \(p = a(1-e^2)\) is the semi-latus rectum, \(e\) is the eccentricity, and \(a\) is the semi-major axis.
When \(e = 1\), the ellipse becomes a parabola, resulting in undefined equations as with our given problem. Understand polar element positioning as:

• Ellipse with \(e < 1\): A more familiar stretched oval.
• \(e = 1\): Can result in a parabola (a special case of ellipse as discussed here).

Students need to check given conditions like the placement of focus to ensure correct subtypes.

Using the Distance Formula in Polar Coordinates

The distance formula in polar coordinates is applied similarly to the Cartesian system, but with focus on radial distances along given angles.
In our scenario, we start with points in polar form, typically represented as \((r, \theta)\). We need to find how far these points are from certain reference points, like the pole.
In the example, given points are:

• Center \(C(3, \frac{\pi}{2})\)
• Vertex \(V(1, \frac{3\pi}{2})\)

To find the distance from the pole to a point in the polar coordinate system, you simply look at the radial component \(r\). For instance, the distance from the pole to vertex \(V\) is simply \(r = 1\). Similarly, from pole to center \(C\), it's \(r = 3\).
In polar systems, always account for the nature of \(r\) and angle \(\theta\) to find distances straightforwardly.