Problem 23 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 19-30: Find an equation for the ellipse that has its center at the origin and satisfies the given conditions. Vertices \(V(0, \pm 6)\), passing through \((3,2)\)

Step-by-Step Solution

Answer

The equation is \( \frac{8x^2}{81} + \frac{y^2}{36} = 1 \).

Step 1: Identify the Ellipse Type

Since the vertices are along the y-axis, the ellipse is vertical.

Step 2: Write Standard Equation Form for Vertical Ellipse

The standard form for a vertical ellipse is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \(b\) is the semi-major axis and \(a\) is the semi-minor axis.

Step 3: Determine Semi-Major Axis Length

The vertices are at \((0, \pm 6)\), so \(b = 6\). Thus, \(b^2 = 36\).

Step 4: Substitute Point into Ellipse Equation

Using the point \((3, 2)\), substitute into \( \frac{x^2}{a^2} + \frac{y^2}{36} = 1 \): \( \frac{3^2}{a^2} + \frac{2^2}{36} = 1 \).

Step 5: Solve for Semi-Minor Axis

Substitute values into the equation: \( \frac{9}{a^2} + \frac{4}{36} = 1 \). Simplify to get \( \frac{9}{a^2} + \frac{1}{9} = 1 \).

Step 6: Clear Fractions and Isolate Variables

Subtract \( \frac{1}{9} \) from both sides: \( \frac{9}{a^2} = \frac{8}{9}\). Then solve for \(a^2\): Multiply by \(9a^2\), giving \(81 = 8a^2\).

Step 7: Calculate Semi-Minor Axis

Divide both sides by 8: \(a^2 = \frac{81}{8}\).

Step 8: Write Final Equation of the Ellipse

Substitute \(a^2\) and \(b^2\) into the ellipse equation: \( \frac{x^2}{\frac{81}{8}} + \frac{y^2}{36} = 1 \). Simplify to \( \frac{8x^2}{81} + \frac{y^2}{36} = 1 \).

Key Concepts

Conic Sections

Conic sections are shapes created by the intersection of a plane with a double-napped cone. Imagine slicing through an ice cream cone at different angles. You can end up with a variety of shapes like circles, ellipses, parabolas, and hyperbolas.
Ellipses are just one of these interesting shapes, and they're defined by their stretched-out rounded appearance. They're like circles that have been squished a bit. This makes them exciting to study because they have unique properties compared to other conic sections.
When working with conic sections, it's crucial to understand if the ellipse is stretched vertically or horizontally, which influences its equation in the standard form. Conic sections provide essential insights into the geometry and algebra of shapes, making them highly valuable in both academic studies and real-world applications.

Vertices and Foci

In ellipses, vertices and foci are vital points that help define their shape and placement. They may sound a bit technical, but they have simple purposes.

• Vertices: These are the points on the ellipse that lie farthest from its center. If you think of an ellipse as an elongated circle, the vertices are at the tips, either side of the elongation.
• Foci: Foci are two crucial points inside the ellipse. The unique property of these points is that, for any point on the ellipse, the sum of the distances from the two foci always remains constant.

Each vertex and focus achieves a specific role in shaping the overall ellipse. For example, in our given exercise, since the vertices are along the y-axis at points (0, ±6), it informs us that the ellipse stretches vertically. This gives the ellipse its unique elongated appearance and helps determine other parameters of the ellipse, such as the equation.

Standard Form of an Ellipse

The standard form of an ellipse's equation provides a mathematical way to describe all the unique geometric properties of an ellipse. The equation changes slightly depending on whether the ellipse is oriented horizontally or vertically.
For a vertical ellipse, the standard form of the equation is given by \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where:

• \(b^2\) is the semi-major axis squared (the longer side of the ellipse).
• \(a^2\) is the semi-minor axis squared (the shorter side of the ellipse).

In the exercise, we determined the vertices at (0, ±6), which means \(b = 6\) and hence \(b^2 = 36\). With a point on the ellipse (3, 2), the task was to find \(a^2\), revealing \(a^2 = \frac{81}{8}\) when placed into the ellipse's equation.
This equation gives us precise control over the shape and orientation of an ellipse, making it a powerful tool for both solving problems and understanding the interaction of geometric shapes.