Problem 51 - Thomas Calculus

Question

Exercises \(49-52\) give equations for ellipses and tell how many units up or down and to the right or left each ellipse is to be shifted. Find an equation for the new ellipse, and find the new foci, vertices, and center. $$ \frac{x^{2}}{3}+\frac{y^{2}}{2}=1, \quad \text { right } 2, \text { up } 3 $$

Step-by-Step Solution

Answer

The new equation is \( \frac{(x-2)^2}{3} + \frac{(y-3)^2}{2} = 1 \). Center: (2,3), Foci: (3,3) and (1,3), Vertices: (\(\sqrt{3}+2, 3\)) and (\(-\sqrt{3}+2, 3\)).

Step 1: Identify the Shift Directions

The given ellipse equation is \( \frac{x^{2}}{3} + \frac{y^{2}}{2} = 1 \). It is stated that the ellipse is shifted right by 2 units and up by 3 units.

Step 2: Write the New Equation

Shifting the ellipse right by 2 units and up by 3 units changes the variables in the equation to \((x-2)\) and \((y-3)\). Substitute these into the original equation: \[ \frac{(x-2)^2}{3} + \frac{(y-3)^2}{2} = 1 \] This is the equation of the new ellipse.

Step 3: Find the New Center

The center of the original ellipse \( \left(0, 0\right) \) is shifted right by 2 units and up by 3 units to become \( \left(2, 3\right) \).

Step 4: Determine the Vertices of the Original Ellipse

For the original ellipse \( \frac{x^{2}}{3} + \frac{y^{2}}{2} = 1 \), the vertices are along the major axis. Since \( b^2 = 3 > a^2 = 2 \), the major axis is horizontal.Vertices: \((\pm \sqrt{3}, 0)\).

Step 5: Calculate the New Vertices

Shift each vertex of the ellipse right by 2 units and up by 3 units:1. Vertex (\(\sqrt{3}, 0\)) becomes (\(\sqrt{3} + 2, 3\)).2. Vertex (\(-\sqrt{3}, 0\)) becomes (\(-\sqrt{3} + 2, 3\)).

Step 6: Determine the Foci of the Original Ellipse

The formula for the foci is \((\pm c, 0)\), where \(c = \sqrt{b^2 - a^2}\). Here, calculate \( c = \sqrt{3 - 2} = 1 \), so the foci are \((\pm 1, 0)\).

Step 7: Calculate the New Foci

Shift each focus of the ellipse right by 2 units and up by 3 units:1. Focus \((1, 0)\) becomes \((1 + 2, 3) = (3, 3)\).2. Focus \((-1, 0)\) becomes \((-1 + 2, 3) = (1, 3)\).

Key Concepts

Ellipse Equation

An ellipse equation is a powerful tool in mathematics that describes the shape and size of an ellipse. It is generally expressed in the standard form for ellipses: \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]Here,

• \( (h, k) \) represents the center of the ellipse, which indicates where the ellipse is centered on the coordinate plane.
• \( a \) and \( b \) are the semi-axes lengths.
• If \( a > b \), the ellipse is stretched along the x-axis, forming a horizontal ellipse. Conversely, if \( b > a \), it is stretched along the y-axis, forming a vertical ellipse.

In our exercise, the original ellipse equation is \( \frac{x^{2}}{3} + \frac{y^{2}}{2} = 1 \). Here, since \( b^2 = 3 \) is greater than \( a^2 = 2\), we have a vertical ellipse. To transform the ellipse, we apply shifts to the variables: a shift to the right changes \(x\) to \(x-2\), and a shift up changes \(y\) to \(y-3\). This alters the equation to: \[ \frac{(x-2)^2}{3} + \frac{(y-3)^2}{2} = 1 \] which represents the new ellipse after the shifts.

Ellipse Vertices

The vertices of an ellipse are crucial elements, marking where the ellipse is widest or tallest. These are the most extended points on each side along either the major or minor axis.

• The major axis is the longest diameter passing through the center and the foci.
• For a vertical ellipse like the one in our exercise, the vertices lie along the major axis, which is the y-axis.

For the original ellipse with equation \( \frac{x^{2}}{3} + \frac{y^{2}}{2} = 1 \), the vertices are located at \((\pm \sqrt{3}, 0)\). After performing the specified shift to the right by 2 units and upward by 3 units, these vertices transform accordingly:

• \( (\sqrt{3}, 0) \) becomes \( (\sqrt{3} + 2, 3) \).
• \( (-\sqrt{3}, 0) \) becomes \( (-\sqrt{3} + 2, 3) \).

These are the vertices of the new ellipse, illustrating how the shape stretches in space.

Ellipse Foci

The foci of an ellipse play a pivotal role in defining its shape and can be found along the major axis. Each focus is a specific point such that the sum of the distances from any point on the ellipse to each focus is constant.Here's how you find the foci:

• Use the formula \( c = \sqrt{b^2 - a^2} \), where \( c \) is the distance from the center to each focus.
• For our exercise with \( \frac{x^{2}}{3} + \frac{y^{2}}{2} = 1 \), substituting \( b^2 = 3 \) and \( a^2 = 2 \), gives \( c = \sqrt{3 - 2} = 1 \).

Thus, the foci of the original ellipse are at \((\pm 1, 0)\). After applying the transformation, these foci move:

• The focus \((1, 0)\) moves to \((1 + 2, 3) = (3, 3)\).
• The focus \((-1, 0)\) shifts to \((-1 + 2, 3) = (1, 3)\).

These are the locations of the foci for the transformed ellipse, crucial for understanding its distance-based geometry.

Ellipse Center

The center of an ellipse is like its balancing point, around which the entire shape is symmetrically oriented. For equations in the standard form, it is defined by the coordinates

• \((h, k)\) in \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \].

In the original ellipse equation \( \frac{x^{2}}{3} + \frac{y^{2}}{2} = 1 \),

• the center is at \((0, 0)\), indicating it is centered at the origin of a coordinate plane.

According to the exercise, it is transformed by moving it right 2 units and up 3 units, leading to a new center at \((2, 3)\). This movement demonstrates how translations affect an ellipse's orientation on a plane, maintaining its shape while relocating its structure fundamentally.