Problem 34 - Algebra and Trigonometry

Question

\(23-34\) Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center, foci, vertices, and lengths of the major and minor axes. If it is a parabola, find the vertex, focus, and directrix. If is a hyperbola, find the center, foci, vertices, and asymptotes. Then sketch the graph of the equation. If the equation has no graph, explain why. $$ x^{2}+4 y^{2}+20 x-40 y+300=0 $$

Step-by-Step Solution

Answer

The equation represents an ellipse with center \((-10, 5)\) and major/minor axes lengths 10/5.

Step 1: Rearrange the Equation

First, we group the like terms of the given equation: \(x^2 + 20x + 4y^2 - 40y + 300 = 0\). Separate the \(x\) and \(y\) terms for completing the square.

Step 2: Completing the Square for X-Terms

Take the \(x\) terms: \(x^2 + 20x\). To complete the square, first, take half of the coefficient of \(x\) (which is 20), getting 10, then square it to get 100. Add and subtract 100 within the \(x\) terms: \((x^2 + 20x + 100) - 100\). This simplifies to \((x + 10)^2 - 100\).

Step 3: Completing the Square for Y-Terms

Now focus on the \(y\) terms: \(4y^2 - 40y\). Factor out the 4: \(4(y^2 - 10y)\). Then complete the square inside the parentheses: half of -10 is -5, squared is 25. Add and subtract 25 in the parentheses: \(4((y^2 - 10y + 25) - 25) = 4((y-5)^2 - 25)\). This simplifies to \(4(y-5)^2 - 100\).

Step 4: Simplify the Equation

Insert the completed squares into the equation and simplify: \((x+10)^2 - 100 + 4(y-5)^2 - 100 + 300 = 0\). Simplifying gives \((x+10)^2 + 4(y-5)^2 = 100\).

Step 5: Identify the Conic

Divide the entire equation by 100 to compare with the standard form: \(\frac{(x+10)^2}{100} + \frac{4(y-5)^2}{100} = 1\). Simplifying yields \(\frac{(x+10)^2}{100} + \frac{(y-5)^2}{25} = 1\). This is an ellipse because there are two positive terms with different denominators.

Step 6: Find the Center, Axes, and Foci

The center of the ellipse is \((-10, 5)\). The semi-major axis length is \(\sqrt{100} = 10\) along the \(x\)-axis, and the semi-minor axis length is \(\sqrt{25} = 5\) along the \(y\)-axis. The foci can be found using the formula \(c^2 = a^2 - b^2\), where \(a = 10\) and \(b = 5\), giving \(c = \sqrt{75} = 5\sqrt{3}\). Thus, the foci are at \((-10 \pm 5\sqrt{3}, 5)\).

Step 7: Sketch the Graph

Draw the ellipse centered at \((-10, 5)\) with vertices at \((-20, 5)\) and \((0, 5)\), and co-vertices at \((-10, 0)\) and \((-10, 10)\). Mark the foci at \((-10 \pm 5\sqrt{3}, 5)\) to complete the sketch.

Key Concepts

Completing the Square

Completing the square is a mathematical technique used to transform a quadratic equation into a special form that makes it easier to analyze, especially for conic sections. This process is essential when dealing with circles, ellipses, parabolas, and hyperbolas.

• First, group the like terms for variables, typically the coefficients of the quadratic terms and the linear terms.
• For each set of terms, such as those involving only the variable x or y, determine half of the coefficient of the linear term, square it, and add and subtract that square within the terms.
• This allows the equation to be re-expressed as a binomial square plus or minus a constant.

Once the terms are rearranged, the equation reveals more information about the shape it represents. In this problem, the terms are completed to form a recognizable ellipse.

Conic Sections

Conic sections are curves obtained by intersecting a right circular cone with a plane. Depending on the angle and location of the intersection, different shapes are obtained: circles, ellipses, parabolas, and hyperbolas.

• An ellipse is formed when the plane cuts across the cone at an angle to the base, but doesn't go through the apex.
• A parabola is a result of a plane parallel to a generator of the cone.
• Hyperbolas arise when the plane intersects both nappes of the cone.

Each conic section can be represented by a specific form of a quadratic equation. In the equation provided, by completing the square, we discovered that it forms an ellipse.

Ellipse Center

The center of an ellipse is the midpoint between the foci and serves as a point from which the ellipse's dimensions are measured. Identifying the center is crucial for understanding and sketching the ellipse.

• For an equation in standard form, the center of the ellipse is represented by the coordinates \(h, k\).
• In the completed ellipse equation \(\frac{(x+10)^2}{100} + \frac{(y-5)^2}{25} = 1\), the center is \((-10, 5)\), which comes from \(x+10 = 0\) and \(y-5 = 0\).

Locating the center helps set a baseline for plotting the rest of the ellipse's features, including its axes and foci.

Ellipse Axes

Ellipses have two types of axes: the major and the minor axes. These axes determine the shape and orientation of the ellipse.

• The major axis is the longest diameter of the ellipse and passes through its center. It corresponds to the higher denominator in the standard equation form.
• The minor axis is perpendicular to the major axis, and is the shortest diameter.
• In \(\frac{(x+10)^2}{100} + \frac{(y-5)^2}{25} = 1\), the lengths of the semi-major and semi-minor axes are determined via the square root of the denominators: \(\sqrt{100} = 10\) for the x-axis (major), and \(\sqrt{25} = 5\) for the y-axis (minor).

Thus, the major axis is 20 units long along the x-direction, while the minor axis is 10 units along the y-direction.

Ellipse Foci

The foci of an ellipse are two distinct points within the ellipse used to define its shape. These points are crucial in understanding the geometry and properties of an ellipse.

• The distance between the center of the ellipse and each focus point is calculated using the formula: \(c = \sqrt{a^2 - b^2}\).
• In our problem, the values are \(a = 10\) and \(b = 5\), leading to \(c = \sqrt{100 - 25} = \sqrt{75} = 5\sqrt{3}\).
• Thus, the foci are located at \((-10 \pm 5\sqrt{3}, 5)\), positioned symmetrically around the center along the major axis.

Finding these points on the graph helps achieve the proper proportions and shape of the ellipse during sketching.