Problem 51 - 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. $$ 9 x^{2}+25 y^{2}-36 x+50 y-164=0 $$

Step-by-Step Solution

Answer

The standard form of the given equation is \((x - 2)^2 / 32.11 + (y + 1)^2 / 11.56 = 1\). The centre of the ellipse is \((2, -1)\), semi-major axis is \(\sqrt{32.11}\), semi-minor axis is \(\sqrt{11.56}\), and the foci are located at \((2 \pm 5.27, -1)\)

Step 1: Regroup the terms

Rearranging the terms to group \(x\) and \(y\) terms, the equation becomes \(9x^2 - 36x + 25y^2 + 50y = 164\).

Step 2: Factoring

Factor out the coefficients of the squared terms from each group to simplify the equation: \(9(x^2 - 4x) + 25(y^2 + 2y) = 164\).

Step 3: Complete the squares

To make each group a perfect square trinomial, add \((b/2)^2\) to each group where \(b\) is the coefficient of the linear term within each group. The equation then becomes \(9[(x - 2)^2 - 4] + 25[(y + 1)^2 - 1] = 164\). After simplifying, it results in \(9(x - 2)^2 + 25(y + 1)^2 = 289\).

Step 4: Convert to Standard Form

Divide every term by 289 on both sides to result in \((x - 2)^2 / (289/9) + (y + 1)^2 / (289/25) = 1\), which simplifies further to \((x - 2)^2 / 32.11 + (y + 1)^2 / 11.56 = 1\)

Step 5: Identify ellipse characteristics

This is a standard form equation of an ellipse. From this, we can find that the centre of the ellipse is \((2, -1)\), the semi-major axis \(a = \sqrt{32.11}\), the semi-minor axis \(b = \sqrt{11.56}\) and the distance of the foci from the centre, \(c\), can be found using the relation \(c=\sqrt{a^{2}-b^{2}}\). After calculating, it results in \(foci = (2 \pm 5.27, -1)\)

Step 6: Graphing the Ellipse

Plot the centre, the vertices, co-vertices, and foci to sketch the ellipse. Lastly, confirm the major and minor axes with the given lengths.

Key Concepts

Completing the Square

Completing the square is a vital algebraic technique used to transform a quadratic expression into a perfect square trinomial, thereby facilitating the solving of quadratic equations. More specifically, it's pivotal when working with the equations of conic sections, including ellipses.

When we have an equation in the form of \(ax^2 + bx + c\), our goal is to create an expression that can be written as \(a(x-h)^2+k\). To do this, we find \((b/2)^2\) and add it to both sides of the equation after which we adjust for the constant. This method is showcased when we rearranged the given elliptic equation and added the necessary constants to each squared term to finalize its conversion into standard form. This skill is not limited to geometry but is also essential in calculus and other higher-level algebra topics.

Standard Form of an Ellipse

The standard form of an ellipse is a neat, organized way of representing the equation of an ellipse, making it much easier to extract crucial information about its graph. For a horizontal ellipse, the standard form is \((x-h)^2/a^2 + (y-k)^2/b^2 = 1\), while for a vertical ellipse, it is \((x-h)^2/b^2 + (y-k)^2/a^2 = 1\), where \((h, k)\) is the centre, \(a\) is the semi-major axis, \(b\) is the semi-minor axis, and the larger denominator corresponds to the squared length of the major axis.

In the given exercise, after completing the square and arranging terms, we achieved this form, which gave us insight into the ellipse's size, orientation, and position. This standard form is a cornerstone for understanding and graphing any ellipse.

Ellipse Foci

The foci of an ellipse are two fixed points located along its major axis, playing a key role in its formal definition and geometric properties. In essence, for any point on the ellipse, the sum of its distances to the foci is constant, and this constant is the length of the major axis.

To find the foci of an ellipse given in standard form, we use the formula \(c = \sqrt{a^2 - b^2}\), where \(c\) is the distance from the centre to each focus, \(a\) is the length of the semi-major axis, and \(b\) is the length of the semi-minor axis. In the provided exercise solution, using this formula revealed the precise locations of the foci relative to the centre of the ellipse.

Conic Sections Algebra

Conic sections are the curves obtained by intersecting a plane with a double-napped cone. These include circles, ellipses, parabolas, and hyperbolas, each represented by specific algebraic equations. The algebra of conic sections involves techniques like completing the square to manipulate an equation into its standard form, allowing us to analyze each curve's attributes and graph it accurately.

Each type of conic section has defining characteristics and equations, and understanding these allows us to solve related mathematical problems and apply them in various science and engineering fields. The exercise provided showcases the algebra of conic sections through completing the square to translate and graph an ellipse - a clear manifestation of the interconnected nature of algebra and geometry in conic sections.