Problem 23 - Algebra and Trigonometry

Question

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 it 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. $$ 4 x^{2}+25 y^{2}-24 x+250 y+561=0 $$

Step-by-Step Solution

Answer

The equation represents an ellipse centered at (3, -5) with vertices at (-2, -5) and (8, -5), and foci at \((3\pm\sqrt{21}, -5)\).

Step 1: Reorganize the Equation

Begin by reorganizing the given equation so that all x terms and y terms are grouped together and the constant is on the right side. The equation is \(4x^2 - 24x + 25y^2 + 250y = -561\).

Step 2: Factor Out the Leading Coefficients

Factor out the coefficients of \(x^2\) and \(y^2\) from their respective groups. This gives us \(4(x^2 - 6x) + 25(y^2 + 10y) = -561\).

Step 3: Complete the Square for x

To complete the square for the \(x\) terms, take half of the coefficient of \(x\) (which is 6), square it to get 9, and add inside the parentheses. Remember to balance by adding \(4 \times 9 = 36\) to both sides. The equation becomes \(4(x^2 - 6x + 9) + 25(y^2 + 10y) = -561 + 36\).

Step 4: Complete the Square for y

For the \(y\) terms, take half of the coefficient of \(y\) (which is 10), square it to get 25, and add inside the parentheses. Balance by adding \(25 \times 25 = 625\) to both sides. After adding, the equation reads \(4(x-3)^2 + 25(y+5)^2 = 100\).

Step 5: Simplify the Equation

Divide the entire equation by 100 to normalize the expression, resulting in \(\frac{(x-3)^2}{25} + \frac{(y+5)^2}{4} = 1\).

Step 6: Identify the Conic Section

The equation is in the form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), which is the standard form for an ellipse. Here, the center is \((3, -5)\), semi-major axis (a) is 5, and semi-minor axis (b) is 2.

Step 7: Calculate Key Features of the Ellipse

The lengths of the major and minor axes are 10 and 4, respectively. The vertices are located at \((3 \pm 5, -5)\) which gives \((-2,-5)\) and \((8,-5)\). The foci can be calculated using \(c = \sqrt{a^2-b^2}\), resulting in \(c = \sqrt{21}\). Hence, the foci are located at approximately \((3 \pm \sqrt{21},-5)\).

Key Concepts

Ellipse

An ellipse is a fascinating shape that looks like a stretched circle. In math, it's a type of conic section, which means it's a shape you get when you slice a cone with a plane.
The standard form of an ellipse equation is:

• \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\)

Here, \((h,k)\) is the center of the ellipse, \(a\) is the length of the semi-major axis (the longest radius), and \(b\) is the length of the semi-minor axis (the shortest radius).
What makes an ellipse special is that for any point along its border, the total distance from two fixed points, called foci, remains constant. Ellipses can be found in many places in real life, like the orbits of planets and the shape of some stadiums.

Complete the Square

Completing the square is a method used to make quadratic expressions into a perfect square trinomial, which then makes it easier to solve equations or convert them into standard forms. When dealing with conic sections like ellipses, it helps us rearrange the equation into a more recognizable form.
In this exercise, for the \(x\) terms, we took half of the linear coefficient, squared it, and added inside the parentheses. The same was done for the \(y\) terms. Completing the square reshapes the equation to expose its real nature, like whether it's an ellipse, parabola, or hyperbola. It's a fundamental tool in algebra that reveals more about the conic section's characteristics.

Major and Minor Axes

The ellipse is defined by its axes: the major axis and the minor axis.
The major axis is the longest diameter of the ellipse, running through the center from vertex to vertex. In our case, the major axis is aligned horizontally with a length of 10 units.
Conversely, the minor axis is the shortest diameter, also passing through the center but perpendicular to the major axis. For this exercise, it is vertical with a length of 4 units. These axes are critical for understanding the overall shape and orientation of the ellipse. Importantly, the major axis always contains the foci, while the minor does not.

Foci

Foci, sometimes called focal points, are two important points inside an ellipse. These points serve a unique purpose: for any point on the ellipse, the sum of the distances to these two foci is constant.
In our example, the foci are calculated as \((3 \pm \sqrt{21},-5)\). We found them by first finding \(c\) using \(c = \sqrt{a^2-b^2}\), where \(a\) and \(b\) are the semi-major and semi-minor axes, respectively.

• Foci give the ellipse its shape and direct its orientation.
• They play a role in real-world phenomena, such as the way sound and light behave in elliptical rooms and devices.

Vertices

Vertices are the points where the ellipse meets its major axis. These are the furthest points on the ellipse from the center, and they help define the ellipse's overall size.
For this example, the vertices are located at \((-2,-5)\) and \((8,-5)\), found by computing \((3 \pm 5, -5)\), based on the center and semi-major axis length.
The vertices indicate the length of the ellipse's longest span. These points are essential when sketching or analyzing the shape of the ellipse and when comparing it to other conic sections, like circles, which have equal radii all around.