Problem 9 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-14: Find the vertices and foci of the ellipse. Sketch its graph, showing the foci. $$ \frac{(x-3)^{2}}{16}+\frac{(y+4)^{2}}{9}=1 $$

Step-by-Step Solution

Answer

Vertices: (7,-4) and (-1,-4); Foci: (3±√7,-4); Center: (3,-4).

Step 1: Identify the Center of the Ellipse

The given equation of the ellipse is \( \frac{(x-3)^{2}}{16} + \frac{(y+4)^{2}}{9} = 1 \). This is in the standard form \( \frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1 \). Here, \( h = 3 \) and \( k = -4 \), so the center of the ellipse is \( (3, -4) \).

Step 2: Determine the Axes Lengths and Orientation

From the equation \( \frac{(x-3)^{2}}{16} + \frac{(y+4)^{2}}{9} = 1 \), we have \( a^2 = 16 \) and \( b^2 = 9 \), giving us \( a = 4 \) and \( b = 3 \). Since \( a > b \), the ellipse is horizontally oriented with the major axis parallel to the x-axis.

Step 3: Find the Vertices

The vertices of the ellipse lie on its major axis, given by the coordinates \( (h \pm a, k) \). Substituting the values, we get the vertices as \( (3 \pm 4, -4) \), so the vertices are \( (7, -4) \) and \( (-1, -4) \).

Step 4: Calculate the Foci

To find the foci, we use the formula for the distance from the center to the foci, \( c = \sqrt{a^2 - b^2} \). Here \( c = \sqrt{16 - 9} = \sqrt{7} \). The foci are located on the major axis at \( (h \pm c, k) \), which gives us the foci \( (3 \pm \sqrt{7}, -4) \). Thus, the foci are \( (3 + \sqrt{7}, -4) \) and \( (3 - \sqrt{7}, -4) \).

Step 5: Sketch the Ellipse

To sketch the ellipse, plot the center \( (3, -4) \), the vertices \( (7, -4) \) and \( (-1, -4) \), and the foci \( (3 + \sqrt{7}, -4) \) and \( (3 - \sqrt{7}, -4) \) on a coordinate plane. Draw an ellipse shape passing through the vertices and centered on \( (3, -4) \), ensuring the major axis is along the direction of the x-axis.

Key Concepts

Vertices of an Ellipse

In the study of ellipses, understanding the vertices is crucial. These are the points where the ellipse makes its widest or tallest reach. Given the standard form of an ellipse's equation, these points are determined by the lengths of the semi-major axis and the ellipse's center.
For the ellipse equation \( \frac{(x-3)^{2}}{16} + \frac{(y+4)^{2}}{9} = 1 \), the center is located at \( (3, -4) \).
The semi-major axis length is \( a = 4 \), indicating that the vertices are horizontal. Hence, the vertices of the ellipse are calculated as \( (h \pm a, k) \).

• Adding \( 4 \): \( (3 + 4, -4) = (7, -4) \)
• Subtracting \( 4 \): \( (3 - 4, -4) = (-1, -4) \)

Therefore, the vertices of the ellipse are \( (7, -4) \) and \( (-1, -4) \). These points lie on the major axis which is horizontally aligned due to \( a > b \).

Foci of an Ellipse

Foci are pivotal as they define the shape of the ellipse and are used in its geometric definition. For an ellipse, foci are two special points located on the major axis.
The equation to find the distance from the center to a focus, \( c \), is \( c = \sqrt{a^2 - b^2} \).
For \( a^2 = 16 \) and \( b^2 = 9 \):

• Calculate \( c: c = \sqrt{16 - 9} = \sqrt{7} \)

The foci positions can be found at \( (h \pm c, k) \):

• Focus 1: \( (3 + \sqrt{7}, -4) \)
• Focus 2: \( (3 - \sqrt{7}, -4) \)

These foci lie inside the ellipse along the major axis, playing a key role in the geometry and the overall equation of the ellipse.

Equation of an Ellipse

The equation of an ellipse can seem complex, but it simplifies when broken down. The standard form is \( \frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1 \), where \( (h,k) \) is the center.
In this form:

• \( a \) represents the semi-major axis if \( a > b \), otherwise, it's the semi-minor axis.
• \( b \) represents the semi-minor axis if \( a > b \), otherwise, it's the semi-major axis.

For the given equation \( \frac{(x-3)^{2}}{16} + \frac{(y+4)^{2}}{9} = 1 \):

• Understanding this, \( h = 3 \) and \( k = -4 \) make the center \( (3, -4) \).
• \( a^{2} = 16 \) thus \( a = 4 \)
• \( b^{2} = 9 \) thus \( b = 3 \)

The major axis is determined to be the x-axis because \( a > b \). This equation guides the shape and orientation of the ellipse.

Major and Minor Axes of an Ellipse

The axes of an ellipse are essential for understanding its geometry. The major axis is the longest diameter, while the minor axis is the shortest one. They intersect at the ellipse's center.
For our ellipse, \( \frac{(x-3)^{2}}{16} + \frac{(y+4)^{2}}{9} = 1 \):

• Major axis has length \( 2a \) with \( a = 4, \) resulting in a total length of \( 8 \).
• Minor axis is \( 2b \) with \( b = 3, \) providing a length of \( 6 \).

Since \( a > b \), the major axis is along the x-axis, running from \( (3 - 4, -4) = (-1, -4) \) to \( (3 + 4, -4) = (7, -4) \).
The minor axis extends from \( (3, -4 - 3) = (3, -7) \) to \( (3, -4 + 3) = (3, -1) \).

Positioning and understanding these axes helps in sketching and comprehending an ellipse's dimensions and orientation.