Problem 43 - Precalculus: Mathematics for Calculus

Question

Finding the Equation of an Ellipse Find an equation for the ellipse that satisfies the given conditions. Foci: \(F(0, \pm \sqrt{10}),\) vertices: \((0, \pm 7)\)

Step-by-Step Solution

Answer

The equation of the ellipse is \( \frac{x^2}{39} + \frac{y^2}{49} = 1 \).

Step 1: Identify the Type of Ellipse

Since the foci and vertices are given along the y-axis, this ellipse is vertical. The standard form of a vertical ellipse is \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \), where \( a \) is the semi-major axis (along the y-axis) and \( b \) is the semi-minor axis (along the x-axis).

Step 2: Determine the Semi-Major Axis, a

The distance from the center of the ellipse (at the origin, \((0,0)\)) to the vertex \((0,7)\) gives us \(a = 7\).

Step 3: Calculate the Distance c Between Center and Foci

The given foci \(F(0, \pm \sqrt{10})\) indicate that \(c = \sqrt{10}\).

Step 4: Use the Relationship c^2 = a^2 - b^2

Use the ellipses' relationship: \( c^2 = a^2 - b^2 \).Substitute the known values:\[ (\sqrt{10})^2 = 7^2 - b^2 \]\[ 10 = 49 - b^2 \]Solving for \(b^2\), \(b^2 = 39\).

Step 5: Write the Equation of the Ellipse

Substitute \(a^2 = 49\) and \(b^2 = 39\) into the equation of the ellipse:\[ \frac{x^2}{39} + \frac{y^2}{49} = 1 \]

Key Concepts

Foci and Vertices

In an ellipse, the foci (focus points) are crucial in defining its shape and orientation. For the given ellipse, the foci are positioned at \( F(0, \pm \sqrt{10}) \), which means they are along the y-axis. The vertices, where the ellipse is widest vertically, are at \((0, \pm 7)\). These points tell us that our ellipse is vertically oriented.

• The foci add up to the idea where the sum of the distances from any point on the ellipse to each focus is constant.
• Vertices indicate the most extended points of the ellipse along its major axis.

In this context, the relationship between the foci and vertices helps us to structure the equation and understand the geometry of an ellipse.

Semi-Major Axis

The semi-major axis of an ellipse is its longest radius. For this vertical ellipse, it's oriented along the y-axis, stretching from the center to the vertex. We determine the length by examining the distance from the center, located at the origin \((0, 0)\), to a vertex at \((0, 7)\).
The length of the semi-major axis, denoted by \(a\), is thus 7, because each vertex is 7 units away from the center along the y-axis.
In an elliptical equation, the square of this length, \(a^2\), appears in the denominator under the \(y^2\) term:\[\frac{y^2}{a^2} \].
This is key to forming the standard equation of an ellipse.

Semi-Minor Axis

The semi-minor axis of an ellipse is the shorter of the two radii, positioned perpendicular to the semi-major axis. For this ellipse, the semi-minor axis lies along the x-axis. To find its length \(b\), we use the relationship among the axes and the foci of any ellipse: \( c^2 = a^2 - b^2 \).
In our example: \[ \text{(Distance to foci)} \ c = \sqrt{10}, \ a = 7 \]
Substituting these into the relationship:\[ (\sqrt{10})^2 = 7^2 - b^2 \]\[ 10 = 49 - b^2 \]
Solving for \(b^2\), we find \(b^2 = 39\).
The actual length of the semi-minor axis \(b\) can then be calculated by taking the square root of 39. Thus, the semi-minor axis dictates the width of the ellipse along the x-axis.

Ellipse Equation

To form the equation of an ellipse, we apply its standard form along with the derived measurements of its semi-major and semi-minor axes.
For a vertical ellipse, the standard equation format is:\[ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \] We substitute \(a^2 = 49\) and \(b^2 = 39\) as determined previously:
\[ \frac{x^2}{39} + \frac{y^2}{49} = 1 \]This equation represents the balance of distances specified for all points lying on the boundary of the ellipse. The denominator values \(a^2\) and \(b^2\) arise from their respective axes and orient the ellipse correctly in the plane with reference to its foci and vertices.
It's crucial for structuring any ellipse's geometry, as they guide how elongated it appears across its axes.