Problem 17 - Calculus with Concepts in Calculus

Question

A triangle is to be inscribed in the ellipse \(\frac{1}{4} x^{2}+y^{2}=1\) with one vertex of the triangle at \((-2,0)\) and the opposite side perpendicular to the \(x\) axis. Find the largest possible area of the triangle.

Step-by-Step Solution

Answer

The largest possible area of the triangle is 2.

Step 1: Identify the ellipse equation

The given equation of the ellipse is \( \frac{1}{4}x^2 + y^2 = 1 \). It can be rewritten in standard form as \( \frac{x^2}{4} + \frac{y^2}{1} = 1 \). This represents an ellipse centered at the origin with semi-major axis along the \(y\)-axis of length 1 and semi-minor axis along the \(x\)-axis of length 2.

Step 2: Establish the triangle parameters

We need to find a triangle inscribed in this ellipse, such that one vertex is at \((-2,0)\). The opposite side should be horizontal (perpendicular to the \(x\)-axis). Therefore, let the coordinates of this side be \((x_1, y_1)\) and \((x_2, y_1)\). Since the side is horizontal, both points lie on the line \(y=y_1\).

Step 3: Relate triangle vertices to ellipse

The points \((x_1, y_1)\) and \((x_2, y_1)\) must satisfy the ellipse equation: \( \frac{x_1^2}{4} + y_1^2 = 1 \) and \( \frac{x_2^2}{4} + y_1^2 = 1 \). By solving these equations, we determine valid \((x_1, y_1)\) and \((x_2, y_1)\) that lie on the ellipse.

Step 4: Calculate area of triangle

The area \(A\) of a triangle with base \((x_1, y_1)\) to \((x_2, y_1)\) and height at \((-2,0)\) is given by \( A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (x_2 - x_1) \times |y_1| \).

Step 5: Find maximum area

To maximize the area, we want to maximize \((x_2 - x_1) \times |y_1|\) subject to \( \frac{x_1^2}{4} + y_1^2 = 1 \) and \( \frac{x_2^2}{4} + y_1^2 = 1 \). Solving for \(x_1\), \(x_2\), and \(y_1\) using calculus or geometric methods, we find that the maximum area occurs when \(y_1 = \pm \frac{1}{\sqrt{2}}\). Inserting this back, the maximum area is \(2\).

Key Concepts

Ellipse Geometry

Ellipses are fascinating geometric shapes that appear in various mathematical and real-world applications. An ellipse is a set of points for which the sum of the distances to two fixed points, called foci, is constant. To describe an ellipse mathematically, we use the standard form equation \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]where:

• \(a\) is the semi-major axis
• \(b\) is the semi-minor axis
• The ellipse is centered at the origin
• If \(a > b\), the ellipse is elongated along the x-axis
• If \(b > a\), the ellipse is elongated along the y-axis

For the equation given in the problem, \(\frac{1}{4}x^2 + y^2 = 1\), we can rewrite it as \(\frac{x^2}{4} + \frac{y^2}{1} = 1\). As a result, it has a semi-major axis of length 2 along the x-axis and a semi-minor axis of length 1 along the y-axis. Understanding these properties is crucial for problems involving shapes inscribed in ellipses.

Triangle Inscribed in Ellipse

When solving problems involving triangles inscribed within ellipses, we need to consider their unique geometric constraints. Imagine a triangle located inside an ellipse, with one vertex fixed at a point like (-2,0). The opposite side is horizontal, meaning it runs parallel to the x-axis. In this scenario, the coordinates of the opposite vertices would be \((x_1, y_1)\) and \((x_2, y_1)\), both positioned along the same horizontal line \(y = y_1\).
These points must satisfy the ellipse equation to be considered within the ellipse, giving us conditions:

• \(\frac{x_1^2}{4} + y_1^2 = 1\)
• \(\frac{x_2^2}{4} + y_1^2 = 1\)

A triangle's base would vary between these two vertices, while the height is the absolute value of the y-coordinate from the fixed point to the line \(y_1\). Solving exercises like this requires effectively utilizing the ellipse's properties to position the triangle's vertices properly.

Maximum Area Problem

In optimization problems like this, we aim to find the maximum possible area of a triangle inscribed in a given shape, such as an ellipse. To determine this, we use calculus or geometric reasoning based on the conditions that the vertices must satisfy. The area of a triangle with a base \(x_2 - x_1\) and height \(|y_1|\) can be expressed as \[ A = \frac{1}{2} \times (x_2 - x_1) \times |y_1| \].
To maximize this area, the product \((x_2 - x_1) \times |y_1|\) needs to be as large as possible while meeting the ellipse conditions. This requires finding suitable \(x_1\), \(x_2\), and \(y_1\) values that maximize this product. Utilizing derivatives or geometric properties, we conclude that the maximum area occurs when the triangle's height is set to a key value, such as \(y_1 = \pm \frac{1}{\sqrt{2}}\). Understanding how each of these parameters interplay leads to solving the maximum area possible, crucial for successfully handling such calculus problems.