Problem 39 - Algebra A Combined Function

Question

Identify whether each equation, when graphed, will be a parabola, circle,ellipse, or hyperbola. Sketch the graph of each equation. If a parabola, label the vertex. If a circle, label the center and note the radius. If an ellipse, label the center. If a hyperbola, label the \(x\) - or \(y\) -intercepts. $$ 9 x^{2}-4 y^{2}=36 $$

Step-by-Step Solution

Answer

The equation is a hyperbola with a center at (0,0) and vertices at (±2,0).

Step 1: Identify the Conic Section

The equation provided is \( 9x^2 - 4y^2 = 36 \). This has the form \( Ax^2 + By^2 = C \), where \( A = 9 \), \( B = -4 \), and \( C = 36 \). Since \( A \) and \( B \) have opposite signs, this equation represents a hyperbola.

Step 2: Rewrite in Standard Form

Divide every term by 36 to begin rewriting the equation in standard form: \[ \frac{9x^2}{36} - \frac{4y^2}{36} = 1 \]. Simplify the fractions to get \[ \frac{x^2}{4} - \frac{y^2}{9} = 1 \]. This is the standard form for a hyperbola with a horizontal transverse axis.

Step 3: Find the Center

The hyperbola \( \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \) is centered at \((h, k)\). In our equation, \( h = 0 \) and \( k = 0 \), so the center is at the origin \((0, 0)\).

Step 4: Determine the Axes and Vertices

For the hyperbola \( \frac{x^2}{4} - \frac{y^2}{9} = 1 \), \( a^2 = 4 \) and \( b^2 = 9 \). Therefore, \( a = 2 \) and \( b = 3 \). The vertices are \( (\pm a, 0) = (\pm 2, 0) \). The distance \( b \) indicates the distance up and down from the center, but it is not part of the vertices.

Step 5: Sketch the Hyperbola

Draw the coordinate axes. Plot the center at \((0, 0)\). Mark the vertices at \((2, 0)\) and \((-2, 0)\). Sketch a rectangle centered at the origin that passes through the points \( (\pm a, \pm b) = (\pm 2, \pm 3) \). Draw the asymptotes passing diagonally through the rectangle corners \( y = \pm \frac{3}{2}x \). Finally, sketch the hyperbola opening horizontally through the vertices.

Key Concepts

Hyperbola

A hyperbola is one of the four major types of conic sections, which are curves formed by the intersection of a plane with a double-napped cone. Unlike ellipses or circles that represent closed curves, a hyperbola consists of two separate, mirror-image open curves. This occurs because in the general form of a hyperbola's equation, the terms involving squares of variables have opposite signs.
When given the equation in the form: \[9x^2 - 4y^2 = 36\]this indicates a hyperbola due to the differing signs on the squared terms. The distinct feature of a hyperbola being its two branches that open either left-right or up-down, helps differentiate it from other conics.
The standard equation of a horizontal hyperbola like \[\frac{x^2}{4} - \frac{y^2}{9} = 1\]is \[\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1,\]where \((h, k)\) is the center, \(a\) is the distance from the center to each vertex along the transverse axis, and \(b\) represents the semi-minor axis. In the equation provided, \(h = 0, k = 0\), indicating that the center is at the origin.

• The transverse axis is horizontal when \(a > b\).
• Each hyperbola's branch approaches, but never intersects, its asymptotes, given by the line equations like \(y = \pm \frac{b}{a}x\).

Equation Graphing

Graphing a hyperbola requires an understanding of its equation's components and transformations. With the equation already in standard form:\[\frac{x^2}{4} - \frac{y^2}{9} = 1\]we identify key elements to sketch correctly.
First, recognize the center at \((0,0)\), the point from which symmetry emanates. Next, calculate \(a = 2\) and \(b = 3\) to anticipate the graph's shape.

• Plot the center on the graph at the origin.
• The vertices will be on the x-axis, at coordinates \((\pm 2, 0)\).
• Construct a rectangle centered at the origin spanning \(\pm a\) horizontally and \(\pm b\) vertically, reaching points \((\pm 2, \pm 3)\).

To show the asymptotic nature of the hyperbola, draw the asymptotes through the rectangle's corners via equations \(y = \pm \frac{3}{2}x\).
These diagonal lines guide the curve as it extends indefinitely along these paths, ensuring that the graph of the hyperbola approaches but never crosses them.Finally, sketch the branches of the hyperbola, opening horizontally, passing through the vertices.

Vertices and Center of Conics

The vertices and center of a conic section are crucial to understand its shape and graph. For hyperbolas, especially, this knowledge helps plot and visualize the curve correctly.
The center of a hyperbola provides the symmetry point for the curve and is the midpoint between the vertices. For our example:
The equation \[\frac{(x-0)^2}{4} - \frac{(y-0)^2}{9} = 1\]indicates a center at the origin \((0, 0)\).

• This center \((h, k)\)—in our equation with \(h = 0\) and \(k = 0\)—is placed at the plane's intersection point.

The vertices of a hyperbola consider the value of \(a\) from the standard form \(\frac{x^2}{a^2}\). They establish where each branch sharply turns, positioned symmetrically around the center.
Here,

• Each vertex is \((\pm a, 0) = (\pm 2, 0)\). This indicates horizontal stretching from the center.

Positioning these elements helps us map and visualize where the hyperbola crosses its principal axis, ensuring a precise graph of the conic.