Problem 69 - Algebra and Trigonometry Real Mathematics, Real People

Question

Show that the polar equation of the ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \quad \text { is } \quad r^{2}=\frac{b^{2}}{1-e^{2} \cos ^{2} \theta}$$

Step-by-Step Solution

Answer

The Cartesian equation of the ellipse can indeed be transformed into the given polar equation. The key was to convert Cartesian into polar coordinates, simplify the resulting equation, and then compare it with the given polar form while considering the eccentricity of an ellipse.

Step 1: Replace \(x\) and \(y\) in the given equation.

Use the relationship between Cartesian and Polar coordinates: \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\). Substituting these into the equation, the resulting equation is \(\frac{r^{2}\cos^{2}(\theta)}{a^{2}} + \frac{r^{2}\sin^{2}(\theta)}{b^{2}} = 1\).

Step 2: Bring the equation to a common denominator.

The common denominator will be \(a^{2}b^{2}\) which yields \(\frac{b^{2}r^{2}\cos^{2}(\theta) + a^{2}r^{2}\sin^{2}(\theta)}{a^{2}b^{2}} = 1\). Now let us rewrite this to match the target form. To rewrite \(sin^{2}(\theta)\) use the Pythagorean identity: \(sin^{2}(\theta)= 1-\cos^{2}(\theta)\). The equation becomes \(\frac{b^{2}r^{2}\cos^{2}(\theta) + a^{2}r^{2}(1-cos^{2}(\theta))}{a^{2}b^{2}} = 1\).

Step 3: Simplification

Simplify the above equation to \(\frac{b^{2}r^{2}\cos^{2}(\theta) + a^{2}r^{2} - a^{2}r^{2}\cos^{2}(\theta)}{a^{2}b^{2}}= 1\). Further simplification leads to \(\frac{(b^{2} - a^{2})r^{2}\cos^{2}(\theta) + a^{2}r^{2}}{a^{2}b^{2}} = 1\). Then, move \(a^{2}b^{2}\) to the other side to obtain the final equation: \((b^{2} - a^{2})r^{2}\cos^{2}(\theta) + a^{2}r^{2} = a^{2}b^{2}\).

Step 4: Identify the Given Form

The given formula in the question, \(r^{2}=\frac{b^{2}}{1-e^{2} \cos ^{2} \theta}\), clearly has the eccentricity of an ellipse \(e = \sqrt{1 - (b^2/a^2)}\). Therefore, square it to make the comparison easier to \(e^2 = 1 - (b^2/a^2)\). Then, solve to end up with \(b^{2} - a^{2} = -a^{2}e^{2}\). Substitute this value in the final equation obtained in step 3 to get \(r^{2}=\frac{b^{2}}{1-e^{2} \cos ^{2} \theta}\).

Key Concepts

Cartesian to Polar Coordinates

Understanding how to convert cartesian coordinates to polar coordinates is vital for solving problems in various branches of mathematics and physics. In cartesian coordinates, we specify the location of a point with an ordered pair \( (x, y) \) based on a two-dimensional plane. To convert to polar coordinates, we use two values: the radius \( r \) and the angle \( \theta \) (theta).

The relationship between cartesian and polar coordinates is given by the equations \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). To find the value of \( r \) and \( \theta \), you can use these formulas: \( r = \sqrt{x^2 + y^2} \) for the radius and \( \theta = \tan^{-1}(\frac{y}{x}) \) for the angle, considering the quadrant where the point lies.

When dealing with an ellipse in polar coordinates, which is the focus of our exercise, we substitute these relationships into the ellipse's standard cartesian equation. The resulting expression often simplifies to a form involving \( r \) as a function of \( \theta \), revealing the elegant shape of an ellipse in polar coordinate form. This is precisely what we do when showing the polar equation of an ellipse.

Eccentricity of an Ellipse

The concept of the eccentricity of an ellipse is a measure of how much the shape of the ellipse deviates from being circular. For a perfect circle, the eccentricity is zero, as the distance from the center to any point on the circumference is constant. However, as the ellipse elongates, the eccentricity grows and has a value between 0 and 1.

The eccentricity \( e \) is defined by the equation \( e = \sqrt{1 - \frac{b^2}{a^2}} \) where \( a \) and \( b \) represent the semi-major and semi-minor axes of the ellipse, respectively. A higher value of eccentricity indicates a more elongated shape. In polar coordinates, the equation of an ellipse can incorporate the eccentricity to express the distance from a focal point to any point on the ellipse as a function of the angle \( \theta \).

This pivotal role of eccentricity is demonstrated in the problem we're dissecting. By identifying the eccentricity within the polar equation, we can understand the geometry of the ellipse more deeply and recognize the interplay between its algebraic expression and geometric form.

Pythagorean Identity

The Pythagorean identity is one of the most important identities in trigonometry, and it's derived from the Pythagorean Theorem which relates the sides of a right triangle. Specifically, the Pythagorean identity states that for any angle \( \theta \), the square of the sine plus the square of the cosine will always be equal to one: \( \sin^2(\theta) + \cos^2(\theta) = 1 \).

This identity is incredibly useful when manipulating trigonometric equations, especially when converting between trigonometric functions as seen in our ellipse problem. By recognizing that \( \sin^2(\theta) \) can be replaced with \( 1 - \cos^2(\theta) \) or vice versa, we can simplify complex expressions and solve problems that involve trigonometric functions of an angle.

In the provided exercise, this identity helps us to rewrite the equation in terms of \( \cos^2(\theta) \) alone, which then allows us to compare it to the given polar equation of the ellipse that incorporates the eccentricity. This step is crucial for demonstrating the stated polar equation and understanding the shape of the ellipse in terms of polar coordinates.