Problem 60 - Algebra and Trigonometry

Question

In Exercises \(51-60,\) convert each equation to standard form by completing the square on \(x\) and \(y .\) Then graph the ellipse and give the location of its foci. $$ 16 x^{2}+25 y^{2}-300 y+500=0 $$

Step-by-Step Solution

Answer

The equation in standard form is \(\frac{x^2}{25} + \frac{(y-6)^2}{20} = 1\). The foci of the ellipse are at (0, 6-\sqrt{5}) and (0, 6+\sqrt{5}).

Step 1: Rewrite the Given Equation

Let's first rewrite the given equation bringing similar terms together. The goal is to make it identifiable as an equation of an ellipse. Rewrite as follows: \(16x^2 + 25(y^2 - 12y + 36) - 900 + 500 = 0\). This allowed us to complete the square for \(y\).

Step 2: Convert the Equation into Standard Form

The standard form for an ellipse centred at the origin, is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Now, rewrite the equation in this form. The result is \(\frac{x^2}{(\sqrt{25})^2} + \frac{(y-6)^2}{(\sqrt{20})^2} = 1\). This equation now reveals that the ellipse is centred at the origin (0,6) with a horizontal radius (the semi-major axis) of \(a = \sqrt{25}=5\) and a vertical radius (the semi-minor axis) of \(b = \sqrt{20} = 2\sqrt{5}\).

Step 3: Locate and identify the foci

To locate the foci of the ellipse, we first must find the distance from the centre of the ellipse to each focus. Use the formula for the distance, \(c = \sqrt{a^2-b^2}\). Calculating this we get \(c = \sqrt{(5)^2-(2\sqrt{5})^2} = \sqrt{25-20} = \sqrt{5}\). Thus, the two foci are located at \((0, 6-\sqrt{5})\) and \((0, 6 + \sqrt{5})\).

Key Concepts

Completing the Square

Completing the square is a powerful method used in algebra to convert quadratic expressions into a form that reveals a lot about their geometric representation. For an ellipse, it serves the purpose of bringing the equation to a recognizable format. The process involves rewriting the quadratic terms related to either the x or y variable as a perfect square trinomial. A trinomial is made perfect when we add and subtract the same value, not to affect the equation's balance.

Let's take the term related to y from the exercise, y^2 - 12y. To complete the square, we take the coefficient of the y term, divide it by 2, and then square it, adding and subtracting this value: y^2 - 12y + 36 - 36. By doing so, we've created a perfect square trinomial, (y-6)^2, which can now be used to express the ellipse in standard form, making it much easier to identify its properties, such as the center and size of its axes.

Ellipse Standard Form

The standard form of an ellipse is essential for understanding its shape and dimensions clearly. It's written as \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) when the ellipse is horizontally oriented, and \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\) when it's vertically oriented. Here, a is the semi-major axis and b is the semi-minor axis of the ellipse.

Once the equation from the exercise is transformed using the completing the square method, we acquire the form \(\frac{x^2}{5^2} + \frac{(y-6)^2}{(2\sqrt{5})^2} = 1\), suggesting that our ellipse is centered not at the origin, but shifted upwards along the y-axis to (0, 6). The values of a and b also indicate the lengths of the axes, allowing us to visualize and graph the shape accurately.

Locating Ellipse Foci

The foci of an ellipse are two fixed points located along the major axis, equidistant from the center, around which the ellipse is shaped. For any point on the ellipse, the sum of the distances to the two foci is constant. To find the foci, we use the formula \(c = \sqrt{a^2-b^2}\) where c is the distance from the center to each focus. The major axis is longer, so it's represented by a, and b denotes the semi-minor axis.

In our case, we have \(c = \sqrt{5}\). Therefore, positioning the foci from our center (0, 6), we move \(\sqrt{5}\) units up and down along the y-axis. This gives us two points: (0, 6 - \sqrt{5}) and (0, 6 + \sqrt{5}), serving as the foci of the ellipse. Knowing the locations of the foci is fundamental for various applications, including optimizing satellite dish placement and understanding planetary orbits.

Conic Sections

Conic sections are the curves obtained by intersecting a plane with a double napped cone. Ellipses are one such shape, formed when the plane cuts through the cone at an angle to its base but not steep enough to produce a hyperbola or parallel to produce a parabola. These shapes have unique algebraic representations and are classified based on their specific equations.

The equation used in this exercise characterizes an ellipse, which is one of the four basic types of conic sections. The others include parabolas, hyperbolas, and circles — each with its own defining features and standard equations. Understanding conic sections and their properties is crucial not only in the study of algebra and geometry but also in various physical applications like the design of telescope lenses, satellite orbits, and even in the paths of planets and stars.