Problem 7 - Precalculus with Calculus Previews

Question

Find the center, foci, vertices, endpoints of the minor axis, and eccentricity of the given ellipse. Graph the ellipse. $$ 9 x^{2}+4 y^{2}=36 $$

Step-by-Step Solution

Answer

The center is (0,0), vertices are (0,3) and (0,-3), endpoints of the minor axis are (2,0) and (-2,0), foci are (0,±√5), and eccentricity is √5/3.

Step 1: Write the given equation in standard form

The given equation is \( 9x^2 + 4y^2 = 36 \). To write it in the standard form of an ellipse, divide the entire equation by 36:\[ \frac{9x^2}{36} + \frac{4y^2}{36} = 1 \]Simplifying, we get:\[ \frac{x^2}{4} + \frac{y^2}{9} = 1 \]

Step 2: Identify the parameters of the ellipse

The standard form of the equation is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Comparing, we have \( a^2 = 4 \) and \( b^2 = 9 \). Thus, \( a = 2 \) and \( b = 3 \). Since \( b > a \), this is a vertically oriented ellipse.

Step 3: Find the center of the ellipse

The center of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) when in standard form \( (x-h)^2/a^2 + (y-k)^2/b^2 = 1 \) is \((h, k)\). Here, it is \((0, 0)\).

Step 4: Determine the vertices

The vertices of a vertically oriented ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) are located at \((h, k \pm b)\). Thus, the vertices are \((0, \pm 3)\), which means they are \((0, 3)\) and \((0, -3)\).

Step 5: Determine the endpoints of the minor axis

The endpoints of the minor axis of a vertically oriented ellipse are \((h \pm a, k)\). Therefore, the endpoints are \((\pm 2, 0)\), which are \((2, 0)\) and \((-2, 0)\).

Step 6: Find the foci

The focal distance \(c\) is found using \( c^2 = b^2 - a^2 \). Thus, \( c^2 = 9 - 4 = 5 \), so \( c = \sqrt{5} \). The foci are located at \((h, k \pm c)\), thus, \((0, \pm \sqrt{5})\), which are approximately \((0, 2.24)\) and \((0, -2.24)\).

Step 7: Calculate the eccentricity

The eccentricity \(e\) is given by \( e = \frac{c}{b} \). With \( c = \sqrt{5} \) and \( b = 3 \), the eccentricity is \( e = \frac{\sqrt{5}}{3} \).

Step 8: Graph the ellipse

To graph the ellipse, plot the center at \((0,0)\), the vertices at \((0, 3)\) and \((0, -3)\), the endpoints of the minor axis at \((2, 0)\) and \((-2, 0)\), and the foci at approximately \((0, 2.24)\) and \((0, -2.24)\). Draw the ellipse shape centered at the origin, stretching vertically.

Key Concepts

Conic Sections

Conic sections are the curves formed by the intersection of a plane with a double-napped cone. When you slice a cone at different angles, you get unique shapes like circles, ellipses, parabolas, and hyperbolas. The ellipse is one of these distinct shapes. Ellipses resemble flattened circles and have unique properties. They are defined algebraically by a quadratic equation and have applications in many fields such as astronomy and physics. In our exercise, the equation for an ellipse was given in a non-standard form and was transformed into a recognizable equation through division by a constant, resulting in \[ \frac{x^2}{4} + \frac{y^2}{9} = 1 \]. Navigating conic sections requires understanding the role of coefficients, which determine the various features and the shape's orientation.

Eccentricity

Eccentricity is a crucial measure in determining how stretched an ellipse is compared to a circle. For conic sections, eccentricity (\( e \)) varies, influencing their shape:

• If \( e = 0 \), the shape is a circle.
• For ellipses, \( 0 < e < 1 \)
• A parabola has \( e = 1 \)
• Hyperbolas have \( e > 1 \)

In ellipses, eccentricity is defined as \( e = \frac{c}{b} \), where \( c \) is the distance from the center to each focus, and \( b \) is the semi-major axis length. In our exercise, after determining the values of \( b \) as 3 and \( c \) as \( \sqrt{5} \), the eccentricity was calculated to be \( \frac{\sqrt{5}}{3} \). This value indicates that the ellipse is slightly elongated but still a bounded curve.

Vertices of Ellipse

The vertices of an ellipse are key points where the ellipse intersects its major axis. These points define the essence of the ellipse's shape and anchor its structure. For a vertically oriented ellipse, the major axis is along the y-axis, with the vertices at positions \((h, k \pm b)\). In our exercise, with the center at \((0,0)\), and given that \( b = 3 \), the vertices are located at \((0, 3)\) and \((0, -3)\). The vertices reflect the maximum distance the ellipse extends in that direction, illustrating the vertical stretch in this scenario. Recognizing these points on a graph helps to accurately sketch the ellipse and understand its orientation.