Problem 1 - Precalculus with Calculus Previews

Question

In Problems 1-20, find the center, foci, vertices, endpoints of the minor axis, and eccentricity of the given ellipse. Graph the ellipse. $$ \frac{x^{2}}{25}+\frac{y^{2}}{9}=1 $$

Step-by-Step Solution

Answer

Center: (0,0); Vertices: (5,0), (-5,0); Minor axis endpoints: (0,3), (0,-3); Foci: (4,0), (-4,0); Eccentricity: 0.8.

Step 1: Identify the Form of the Equation

The equation given is \( \frac{x^2}{25} + \frac{y^2}{9} = 1 \), which is in the standard form of an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Here, \( a^2 = 25 \) and \( b^2 = 9 \). Since \( a^2 > b^2 \), this is a horizontal ellipse.

Step 2: Determine the Center of the Ellipse

The center of an ellipse in the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is at the origin \((0,0)\), since there are no \(h\) or \(k\) shifts added to the equation.

Step 3: Find the Vertices

For a horizontal ellipse, the vertices are \((\pm a, 0)\). Here, \(a = \sqrt{25} = 5\). So, the vertices are \((5, 0)\) and \((-5, 0)\).

Step 4: Locate the Endpoints of the Minor Axis

The minor axis endpoints for a horizontal ellipse are \((0, \pm b)\). Here, \(b = \sqrt{9} = 3\). Thus, the endpoints are \((0, 3)\) and \((0, -3)\).

Step 5: Calculate the Foci

The foci position for a horizontal ellipse is \((\pm c, 0)\) where \(c\) is calculated using \(c^2 = a^2 - b^2\). Here, \(c^2 = 25 - 9 = 16\) so \(c = \sqrt{16} = 4\). Therefore, the foci are \((4, 0)\) and \((-4, 0)\).

Step 6: Determine the Eccentricity

The eccentricity \(e\) is given by the formula \(e = \frac{c}{a}\). With \(c = 4\) and \(a = 5\), we get \(e = \frac{4}{5} = 0.8\).

Step 7: Graph the Ellipse

To graph the ellipse, plot the center at \((0,0)\), the vertices at \((5,0)\) and \((-5,0)\), the endpoints of the minor axis at \((0,3)\) and \((0,-3)\), and the foci at \((4,0)\) and \((-4,0)\). Connect these points with an elliptical shape.

Key Concepts

Center of Ellipse

The center of an ellipse plays a crucial role in its geometry. It's the midpoint from which all measurements in the ellipse are taken. For the standard equation of an ellipse, which looks like \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), the center is at the origin, \( (0,0) \), unless otherwise shifted by additional terms \( h \) and \( k \) in the general form. Understanding the center is important because it defines the precise location on a coordinate plane where the ellipse is balanced. Many mathematical concepts of the ellipse like symmetry and the location of other key points such as vertices and foci revolve around the center.

Vertices of Ellipse

The vertices of an ellipse are the points where the ellipse is widest or tallest, depending on its orientation. These points are essential because they define the major axis' reach. For our ellipse, \( \frac{x^2}{25} + \frac{y^2}{9} = 1 \), it is aligned horizontally since \( a^2 > b^2 \).

• The vertices are positioned at \( (\pm a, 0) \).
• Here, \( a = \sqrt{25} = 5 \).
• This gives us the specific vertices \( (5, 0) \) and \( (-5, 0) \).

The vertices not only help to sketch the ellipse but also to calculate other vital components like the foci and eccentricity.

Foci of Ellipse

The foci (plural of focus) of an ellipse are two points inside the ellipse used to define its shape. The sum of the distances from any point on the ellipse to the foci is constant.To find the foci:

• We use the formula \( c^2 = a^2 - b^2 \).
• For our equation, \( c^2 = 25 - 9 = 16 \).
• Thus, \( c = \sqrt{16} = 4 \).
• The foci are located at \( (\pm c, 0) \) which means \( (4, 0) \) and \( (-4, 0) \).

The foci are pivotal in defining the eccentricity and the overall geometry of the ellipse.

Eccentricity of Ellipse

Eccentricity measures how "stretched" an ellipse is. It is a ratio, describing how much the ellipse deviates from being a circle (which has an eccentricity of 0).

• The formula for finding the eccentricity \( e \) is \( e = \frac{c}{a} \).
• For our problem, since \( c = 4 \) and \( a = 5 \), \( e = \frac{4}{5} = 0.8 \).

An eccentricity value less than 1 confirms that the shape is indeed an ellipse. In this case, an \( e \) of 0.8 suggests a fairly stretched elliptical shape. Understanding eccentricity is essential for analyzing the tightness or flatness of the ellipse's curve.