Problem 14 - Algebra for College Students

Question

Find the vertices, the endpoints of the minor axis, and the foci of the given ellipse, and sketch its graph. See answer section. $$ \frac{(x+3)^{2}}{16}+\frac{(y-2)^{2}}{4}=1 $$

Step-by-Step Solution

Answer

Vertices are (-7, 2) and (1, 2); endpoints of minor axis are (-3, 0) and (-3, 4); foci are (-3 - 2√3, 2) and (-3 + 2√3, 2).

Step 1: Identify Standard Form

The given equation is \( \frac{(x+3)^2}{16} + \frac{(y-2)^2}{4} = 1 \). This is in the standard form of an ellipse \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), where \( (h, k) \) is the center, \( a^2 \) is the larger denominator, and \( b^2 \) is the smaller denominator.

Step 2: Determine Ellipse Orientation

Since \( a^2 = 16 > 4 = b^2 \), the ellipse is horizontal. This means the major axis is along the x-direction with \( a = \sqrt{16} = 4 \) and the minor axis is vertical with \( b = \sqrt{4} = 2 \).

Step 3: Find the Center

The center \( (h, k) \) of the ellipse is determined from the transformation of coordinates: \((x+3)^2\) means \(h = -3\) and \((y-2)^2\) means \(k = 2\). Thus, the center is \((-3, 2)\).

Step 4: Calculate Vertices

The vertices of a horizontal ellipse are located \( a \) units left and right of the center. So, the vertices are \((-3-4, 2)\) and \((-3+4, 2)\), which simplifies to \((-7, 2)\) and \((1, 2)\).

Step 5: Determine Endpoints of Minor Axis

The endpoints of the minor axis are located \( b \) units up and down from the center. So, the endpoints are \((-3, 2-2)\) and \((-3, 2+2)\), which simplifies to \((-3, 0)\) and \((-3, 4)\).

Step 6: Find the Foci

For a horizontal ellipse, the distance to each focus from the center is calculated using \( c = \sqrt{a^2 - b^2} = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3} \). Thus, the foci are located at \((-3 - 2\sqrt{3}, 2)\) and \((-3 + 2\sqrt{3}, 2)\).

Step 7: Sketch the Graph

To sketch the ellipse, plot the center at \((-3, 2)\), draw the major axis horizontally from \((-7, 2)\) to \((1, 2)\), plot the minor axis endpoints \((-3, 0)\) and \((-3, 4)\), and indicate the foci at \((-3 - 2\sqrt{3}, 2)\) and \((-3 + 2\sqrt{3}, 2)\). Finally, sketch the ellipse by connecting the vertices softly in an oval shape.

Key Concepts

Vertices of an Ellipse

The vertices of an ellipse are crucial points that lie on the major axis of the ellipse. To find the vertices, you'll need to know the center and the length of the semi-major axis, labeled as "a".

In the equation \( \frac{(x+3)^2}{16} + \frac{(y-2)^2}{4} = 1 \), we see that the center of the ellipse is \((-3, 2)\). The term under \( (x+3)^2 \) is 16, indicating the square of the semi-major axis, so \( a = \sqrt{16} = 4 \). Since the ellipse is horizontal, the semi-major axis stretches along the x-axis.

The vertices are calculated by moving "a" units right and left from the center. This calculation gives us the vertices at \((-3-4, 2)\) and \((-3+4, 2)\), which simplifies to \((-7, 2)\) and \((1, 2)\). These points mark the widest span of the ellipse.

Foci of an Ellipse

The foci are another set of significant points inside an ellipse. Understanding them is crucial as they determine the ellipse's shape and uniqueness.

To calculate the foci, use the formula \( c = \sqrt{a^2 - b^2} \), where "c" is the distance from the center to each focus. For this ellipse, \( a^2 = 16 \) and \( b^2 = 4 \), so \( c = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3} \).

Since our ellipse is horizontal, the foci are aligned along the x-axis, horizontally spaced. Therefore, each focus is \( 2\sqrt{3} \) units left and right of the center, giving locations \((-3 - 2\sqrt{3}, 2)\) and \((-3 + 2\sqrt{3}, 2)\). These points are essential for understanding how light or energy will behave within the ellipse.

Standard Form of an Ellipse

Ellipses have a standard equation form that aids in identifying their properties quickly. The standard form is given by \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), where \((h, k)\) represents the center of the ellipse.

For the equation \( \frac{(x+3)^2}{16} + \frac{(y-2)^2}{4} = 1 \), we identify \((h, k)\) as \((-3, 2)\), determining the ellipse's center. "a" and "b" denote semi-major and semi-minor axes, recognizable by their squares in the denominators. Here, \( a^2 = 16 \) and \( b^2 = 4 \), thus \( a = 4 \) and \( b = 2 \).

The larger denominator (16) belongs to the term \((x+3)^2\), implying a horizontal major axis as it extends along the x-axis. Recognizing this form allows you to pinpoint the center, axis lengths, and orientation of the ellipse at a glance.