Problem 37 - Algebra 2

Question

Find the foci for each equation of an ellipse. $$ 36 x^{2}+8 y^{2}=288 $$

Step-by-Step Solution

Answer

The foci of the given ellipse are located at \( (0, \pm \sqrt{28}) \).

Step 1: rewrite the equation in standard form

To rewrite the equation \(36x^2 + 8y^2 = 288\) in standard form, both sides of the equation need to be divided by 288, resulting in the equation \(\frac{x^2}{8} + \frac{y^2}{36} = 1\). Now, the ellipse is in the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), with \(a^2 = 8\) and \(b^2 = 36\).

Step 2: find the semi-major and semi-minor axes

Since \(a^2 = 8\), we can find that \(a = \sqrt{8}\), and since \(b^2 = 36\), we can find that \(b = \sqrt{36} = 6\). The semi-major axis \(a\) is smaller than the semi-minor axis \(b\). In this case, the foci of the ellipse will be along the y-axis, as the larger value is under \(y^2\).

Step 3: calculate the foci

The foci of the ellipse can be calculated using the formula \(foci = \pm c\), where \(c = \sqrt{b^2 - a^2}\), because \(b > a\). Substituting \(a = \sqrt{8}\) and \(b = 6\), the calculation becomes \(c = \sqrt{36 - 8} = \sqrt{28}\). Therefore, the foci of the ellipse are located at \(0,\pm \sqrt{28}\).

Key Concepts

Ellipse Equation

Ellipses are fascinating curves in geometry, forming an elongated circle or oval. The general form of an ellipse equation gives us the structure we need to explore its properties. For any ellipse centered at the origin, the equation typically looks like: \[ Ax^2 + By^2 = C \] where \(A\), \(B\), and \(C\) are constants. This equation might initially seem daunting! So why does it matter? This equation gives us a starting point – it tells us what the ellipse will look like and allows us to determine other important features, like axes and foci.
For instance, the given problem states an ellipse equation as \(36x^2 + 8y^2 = 288\). To better understand and visualize this ellipse, we need to transform it into something more recognizable. That's where the standard form comes in!

Standard Form of Ellipse

To understand an ellipse, converting its equation to the standard form is crucial. The standard form we aim for is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] This form makes it easier to identify the ellipse's axes lengths and other properties.

• \(a^2\) and \(b^2\) are the squares of the semi-major and semi-minor axes.
• The value beneath \(x^2\) is the square of the semi-major axis if \(a^2 \geq b^2\), otherwise, it belongs to the semi-minor axis.

To convert the original equation \(36x^2+8y^2=288\) into this easier-to-read form, divide every term by 288 to simplify:
\[ \frac{x^2}{8} + \frac{y^2}{36} = 1 \] Now, it's clear which terms relate to which axes. The largest denominator corresponds to the semi-major axis, helping us go a step further in understanding the ellipse's dimensions.

Semi-Major Axis

The semi-major axis is a fundamental part of an ellipse's structure. In an ellipse, the semi-major axis is the longest distance from the center to the edge, defining its widest point along one principal direction.
In our transformed equation \(\frac{x^2}{8} + \frac{y^2}{36} = 1\), the denominator \(36\) belongs to the term \(y^2\). Hence, the semi-major axis is associated with the y-direction, revealing an elongated shape vertically. To find its length, calculate the square root of \(b^2\):

• \(b^2 = 36 \Rightarrow b = \sqrt{36} = 6\)

Thus, the semi-major axis length is 6 units, determining how much longer this axis is than its counterpart. This information is essential for determining the foci, which are key identifiers of an ellipse's geometry.

Semi-Minor Axis

Equally significant to an ellipse's structure is its semi-minor axis. This axis spans the shortest distance from the center to the ellipse's perimeter, perpendicular to the semi-major axis. Although shorter, it’s just as vital to completing the ellipse's oval form.
From our equation \(\frac{x^2}{8} + \frac{y^2}{36} = 1\), the smallest denominator \(8\) associates with \(x^2\). Thus, the semi-minor axis runs horizontally, less elongated compared to its vertical counterpart. To determine its length, take the square root of \(a^2\):

• \(a^2 = 8 \Rightarrow a = \sqrt{8} \approx 2.83\)

With this measurement, you can comprehend the full shape and orientation of the ellipse. Knowing the axis lengths allows us to swiftly find the foci, providing a complete picture of its geometric layout.