Problem 21 - A Graphical Approach to College Algebra

Question

Graph each ellipse by hand. Give the domain and range. Give the foci and identify the center. Do not use a calculator. $$\frac{(x+1)^{2}}{64}+\frac{(y-2)^{2}}{49}=1$$

Step-by-Step Solution

Answer

Center: \((-1, 2)\); Foci: \((-1\pm\sqrt{15}, 2)\); Domain: \([-9, 7]\); Range: \([-5, 9]\).

Step 1: Identify Center

The standard form of an ellipse is \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \). For the given equation \( \frac{(x+1)^2}{64} + \frac{(y-2)^2}{49} = 1 \), identify \( h \) and \( k \). Here, \( h = -1 \) and \( k = 2 \), so the center of the ellipse is \((-1, 2)\).

Step 2: Determine Vertices

In the equation \( \frac{(x+1)^2}{64} + \frac{(y-2)^2}{49} = 1 \), \( a^2 = 64 \) and \( b^2 = 49 \). Since \( a^2 > b^2 \), the major axis is horizontal. The vertices are at \((-1-a, 2)\), \((-1+a, 2)\), where \( a = \sqrt{64} = 8 \). Thus, the vertices are \((-9, 2)\) and \((7, 2)\).

Step 3: Determine Co-Vertices

The co-vertices lie on the minor axis. For the given equation, \( b = \sqrt{49} = 7 \). The co-vertices are at \((-1, 2-b)\) and \((-1, 2+b)\), which are \((-1, -5)\) and \((-1, 9)\).

Step 4: Identify Foci

The foci of the ellipse are calculated using \( c = \sqrt{a^2 - b^2} \). Here, \( a^2 = 64 \) and \( b^2 = 49 \), so \( c = \sqrt{64 - 49} = \sqrt{15} \). The foci are at \((-1-c, 2)\) and \((-1+c, 2)\), which are approximately \((-1-\sqrt{15}, 2)\) and \((-1+\sqrt{15}, 2)\).

Step 5: State Domain and Range

The domain of the ellipse, determined by the horizontal major axis, ranges from the leftmost to the rightmost vertex: \([-9, 7]\). The range, determined by the vertical extent of the ellipse, is from the bottom to top co-vertex: \([-5, 9]\).

Step 6: Graph the Ellipse

Draw the center at \((-1, 2)\), plot the vertices \((-9, 2)\) and \((7, 2)\), and the co-vertices \((-1, -5)\) and \((-1, 9)\). Sketch the ellipse by connecting these points in an elongated oval shape. Remember, the ellipse is wider horizontally due to the major axis being horizontal.

Key Concepts

Graphing Ellipses

The first step in graphing an ellipse is understanding its standard form equation: \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \).
This classic formula is crucial for identifying the key elements of the ellipse. Here, \( h \) and \( k \) represent the x and y coordinates of the center, respectively, while \( a^2 \) and \( b^2 \) are the squares of the semi-major and semi-minor axes.

To graph the ellipse, start by plotting its center. Then, determine the vertices along the major axis and co-vertices along the minor axis.
The final graph is an oval shape that touches all these points.

In our specific example, given the equation \( \frac{(x+1)^2}{64} + \frac{(y-2)^2}{49} = 1 \), notice that \( a^2 = 64 \) and \( b^2 = 49 \). This tells us that the semi-major axis is horizontal because \( a^2 > b^2 \).
Here, you would draw an oval that is longer horizontally than vertically.

Center of Ellipse

The center of an ellipse is a vital point, as it is the midpoint of its major and minor axes, making it the most balanced point of the shape.
This concept directly informs the ellipse's position on the graph.

For our equation, \( \frac{(x+1)^2}{64} + \frac{(y-2)^2}{49} = 1 \), the center is found by identifying the values \( h \) and \( k \) from the equation.
Here, \( h = -1 \) and \( k = 2 \), so the center is at \((-1, 2)\). This means the center is slightly to the left and slightly above the origin on a Cartesian coordinate system.

Plotting this point on your graph is your starting mark for plotting an ellipse.

Foci of Ellipse

Foci are unique and critical points within an ellipse. They are used to define the precise geometric properties of the shape. The concept of foci is particularly significant because for any point on the ellipse, the sum of the distances to the foci is constant.

To find the foci, we use the formula \( c = \sqrt{a^2 - b^2} \), which calculates the distance from the center to each focus.
For our equation, with \( a^2 = 64 \) and \( b^2 = 49 \), then \( c = \sqrt{64 - 49} = \sqrt{15} \).
Thus, the foci are at approximately \((-1-\sqrt{15}, 2)\) and \((-1+\sqrt{15}, 2)\), indicating their positions along the major axis.
Understanding where the foci are helps in structurally forming the ellipse precisely.

Domain and Range of Ellipse

Determining the domain and range of an ellipse informs us about the extent of the ellipse horizontally and vertically.
The domain spans from the leftmost to the rightmost points, and the range spans from the lowest to the highest points.

For our ellipse, the vertices and the co-vertices tell us these bounds. With vertices at \((-9, 2)\) and \((7, 2)\), the domain is \([-9, 7]\).
This range covers the broad span of the ellipse from left to right. Similarly, with co-vertices at \((-1, -5)\) and \((-1, 9)\), the range is \([-5, 9]\).

Knowing the domain and range helps in drawing the boundary of the ellipse, ensuring it's graphed accurately within these limits.