Problem 23 - Algebra 2

Question

Identify the conic section represented by each equation by writing the equation in standard form. For a parabola, give the vertex. For a circle, give the center and the radius. For an ellipse or a hyperbola, give the center and the foci. Sketch the graph. \(9 x^{2}-4 y^{2}-24 y=72\)

Step-by-Step Solution

Answer

The given conic equation represents a hyperbola with center at (0, -3) and foci at (0, -3 + sqrt(10)) and (0, -3 - sqrt(10)).

Step 1: Convert to Standard Form

First step is to convert the equation into standard form. The given equation is \(9 x^{2}-4 y^{2}-24 y=72\). Rewrite it as \(9x^2 - 4(y^2 + 6y - 9) = 72\) and simplify it to get \(x^2-\frac{4}{9}(y+3)^2=8\)

Step 2: Identify the Conic Section

After converting the equation into standard form, we can observe that this equation has both a positive \(x^2\) term and a negative \(y^2\) term. Therefore this equation represents a hyperbola.

Step 3: Identify Center and Foci of Hyperbola

For a hyperbola equation in the form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the center of the hyperbola is \((h, k)\). The foci are at \((h \pm \sqrt{a^2+b^2}, k)\). Here \(h = 0\), \(k = -3\), \(a = \sqrt{8}\) and \(b = \sqrt{2}\). So, the center is at (0, -3) and the foci are at \((0, -3 \pm \sqrt{10})\).

Step 4: Sketch the Graph

To sketch the graph, plot the center on an xy-plane. Then using the value of a, draw two horizontal lines tangential to the vertex - these are the guide lines for sketching the hyperbola. Also plot the foci. Finally sketch the hyperbola using the guide lines with branches opening along the x-axis.

Key Concepts

Conic Sections

A conic section is a curve obtained by intersecting a cone with a plane. Depending on the angle and position of the intersecting plane, we can get different types of curves, including circles, ellipses, parabolas, and hyperbolas. These curves are ubiquitous in both nature and mathematics and play crucial roles in various fields such as astronomy, physics, and engineering.

Each type of conic section has unique characteristics:

• Circle: All points equidistant from a center point.
• Ellipse: The sum of distances from two foci to any point on the ellipse is constant.
• Parabola: Any point on it is equidistant from a focus and a directrix.
• Hyperbola: The difference of distances from two foci to any point on the hyperbola is constant.

In the exercise you encountered, finding which conic section is represented comes down to analyzing the given equation.

Standard Form of Hyperbolas

Understanding the standard form of hyperbolas is critical in identifying and analyzing them. A hyperbola can be represented by its standard equation, which looks similar to an ellipse but with one crucial difference – the difference between the squares of the variables. This can seem a bit tricky but stick with it!The standard form of the hyperbola's equation usually is:

• extbf{Horizontal Hyperbola:} displayed as displayed as \(\frac{{(x-h)^2}}{a^2} - \frac{{(y-k)^2}}{b^2} = 1\)
• extbf{Vertical Hyperbola:} \(\frac{{(y-k)^2}}{a^2} - \frac{{(x-h)^2}}{b^2} = 1\)

In these equations, \(h\) and \(k\) define the center of the hyperbola. Meanwhile, \(a\) and \(b\) dictate how far the vertices and the asymptotes stray from this center, respectively.For the exercise, you converted your equation into a standard form and determined how these values relate to the hyperbola’s orientation.

Equations of Hyperbolas

The equations of hyperbolas consist of two squares, similar to those of ellipses. However, in hyperbolas, one is subtracted from the other, which results in their unique "opening" characteristic.In the context of your exercise, you started with the equation\(9x^2 - 4y^2 - 24y = 72\). Transforming this into standard form is an essential step. You do this by rearranging and simplifying it into \(x^2 - \frac{4}{9}(y + 3)^2 = 8\). During this transformation, notice the presence of both a positive and a negative term:

• The presence of both positive \(x^2\) and negative \(y^2\) indicates we are dealing with a hyperbola.
• By factoring and rearranging terms, it’s now easier to identify key components like the hyperbola's center and foci.

This approach sets the stage for sketching its graph and understanding the hyperbola's configuration in space.

Graphing Hyperbolas

Graphing hyperbolas may sound intimidating initially, but following a methodical approach simplifies the process. In your exercise, after identifying your hyperbola's standard form, you can start sketching!First, plot the hyperbola's center on the coordinate plane, which serves as your focal point. For the given example, your center was at \((0, -3)\). From here, it's helpful to plot the foci by using their calculated positions, \((0, -3 \pm \sqrt{10})\).The values of \(a\) and \(b\) help you sketch the asymptotes, imaginary lines that guide the hyperbola's branches. The branches, or the curves themselves, will "open" in the direction of the variable with the positive coefficient, either along the x-axis or y-axis, depending on the standard form used.Your hyperbola will have vertices placed at a distance of \(a\) from the center but along the axis of symmetry. Once you've plotted these key points, use the asymptotes to guide the curvature of each branch, allowing them to "approach" without ever touching these lines.