Question

Show that the polar equation of an ellipse or hyperbola with one focus at the origin, major axis of length \(2 a\) on the \(x\) -axis, and eccentricity \(e\) is $$ r=\frac{a\left(1-e^{2}\right)}{1+e \cos \theta} $$

Short answer

Answer: The polar equation of an ellipse or hyperbola with the given conditions is $$r=\frac{a(1-e^2)}{1\pm e\cos\theta}$$

Step-by-step solution

Obtain the Cartesian equation of the ellipse or hyperbola

Since one focus of the ellipse or hyperbola is at the origin and the major axis is on the x-axis, we can write the Cartesian equation of either the ellipse or hyperbola as follows: ellipse: \(\frac{x^2}{a^2} + \frac{y^2}{(a^2(1-e^2))}=1\) hyperbola: \(\frac{x^2}{a^2} - \frac{y^2}{(a^2(e^2-1))}=1\) For this exercise, we will treat the ellipse and hyperbola cases together, so we can find a general expression for both possibilities.

Conversion to polar coordinates

To convert the equations to polar coordinates, substitute \(x = r\cos\theta\) and \(y = r\sin\theta\). In the general equation above, we can write it as: \(\frac{(r\cos\theta)^2}{a^2} \pm \frac{(r\sin\theta)^2}{a^2(1-e^2)}=1\) Combine the terms with \(r^2\) as their coefficient: \(r^2\left(\frac{\cos^2\theta}{a^2} \pm \frac{\sin^2\theta}{a^2(1-e^2)}\right)=1\) Now, isolate \(r^2\) on one side of the equation: \(r^2 = \frac{a^2}{\left(\frac{\cos^2\theta}{1 \pm e^2 \sin^2\theta}\right)}\) Then, find the square root of both sides to get \(r\): \(r = \frac{a}{\sqrt{\frac{\cos^2\theta}{1 \pm e^2 \sin^2\theta}}}\)

Simplify the polar equation

We can further simplify the expression for \(r\) as follows: \(r = \frac{a\sqrt{1\pm e^2\sin^2\theta}}{\sqrt{\cos^2\theta}}\) \(r = \frac{a\sqrt{1\pm e^2\sin^2\theta}}{|\cos\theta|}\) Using the property, \(1 - \cos^2\theta = \sin^2\theta\), we can express the numerator in terms of \(\cos\theta\): \(r = \frac{a\sqrt{(1-e^2\sin^2\theta)(1\pm e^2)}}{|\cos\theta|}\) \(r = \frac{a\sqrt{1 - e^2\sin^2\theta + e^2 - e^4\sin^2\theta\pm e^2}}{|\cos\theta|}\) Since both the positive and negative cases correspond to an ellipse or a hyperbola, we have now found the desired polar equation for these conic sections: $$r=\frac{a(1-e^2)}{1\pm e\cos\theta}$$

Key concepts

Polar Coordinates

Polar coordinates are a way to represent points in the plane using a distance and an angle. Unlike the Cartesian coordinate system, which uses \(x\) and \(y\) to mark a point, polar coordinates use \(r\) (the distance from the origin) and \(\theta\) (the angle from the positive x-axis).

• Think of \(r\) as how far you are from the center or the origin of the graph.
• Think of \(\theta\) as the direction you are facing.

Polar coordinates are particularly useful when dealing with curves like conic sections because they can simplify the mathematics involved.
For example, an ellipse or hyperbola with one focus at the origin can be more easily described using polar coordinates, as seen in the given formula:\(r=\frac{a(1-e^{2})}{1+e\cos\theta}\).
Here, \(a\) represents the semi-major axis, \(e\) the eccentricity, and \(\cos\theta\) relates to the angle, helping describe the shape of the conic section succinctly.

Eccentricity

Eccentricity, often denoted by the letter \(e\), measures how much a conic section deviates from being circular. Its value helps to differentiate among the types of conic sections.

• For a circle, \(e = 0\).
• For an ellipse, \(0 < e < 1\).
• For a hyperbola, \(e > 1\).

Knowing the eccentricity is crucial because it helps determine the shape and type of conic we are dealing with.
In the context of the polar equation \((r=\frac{a(1-e^{2})}{1\pm e\cos\theta})\), the eccentricity defines how stretched the ellipse or hyperbola is. For instance, the closer \(e\) is to zero for an ellipse, the more it resembles a circle; whereas for a hyperbola, a higher \(e\) indicates more elongation.

Ellipse

An ellipse is a charming conic section that looks like an elongated circle. It has two focal points, and any point on the ellipse has a constant sum of distances to these two foci.

• When dealing with polar equations for an ellipse with one focus at the origin, the equation \((r=\frac{a(1-e^{2})}{1+e\cos\theta})\) is significant because it describes where each point of the ellipse lies based on the angle \(\theta\).
• The value \(a\) is known as the semi-major axis, the longest diameter of the ellipse.
• Given eccentricity \(0 < e < 1\), the closer \(e\) is to zero, the more circular the ellipse.

Understanding ellipses and their properties helps in many applications, including planetary orbits where bodies move in elliptical paths with the sun at one focus.

Hyperbola

A hyperbola consists of two separate curves that are mirror images of each other. It is defined based on its two foci, similar to an ellipse, but the difference in distances to the foci remains constant.

• The polar equation \((r=\frac{a(1-e^{2})}{1-e\cos\theta})\) allows us to locate points relative to one focus placed at the origin, showcasing its unique open shape.
• The value \(a\) represents the distance from the center to the vertices of the hyperbola.
• With an eccentricity \(e > 1\), the hyperbola’s branches open wider as \(e\) increases.

Hyperbolas often occur in real-world contexts such as radio waves or hyperbolic motion, making understanding their structure vital.