Problem 30 - Precalculus

Question

Find the standard form of the equation of the ellipse with the given characteristics. $$\text { Center: }(1,-5) ; a=5 c ; \text { foci: }(1,-6),(1,-4)$$

Step-by-Step Solution

Answer

The standard form of the equation of the ellipse is \(\frac{(x-1)^2}{5^2} + \frac{(y+5)^2}{24} = 1\).

Step 1: Determine Orientation and Calculate a

Given the foci at (1,-6) and (1,-4), this suggests that the ellipse is vertically oriented since the foci have the same x-value but different y-values. Also, from the given relation between a and c, we have \(a=5c\).

Step 2: Calculate c

The value of c which is the distance from the center to a foci can be calculated from the given center (1,-5) and one of the foci (1,-4). Therefore, \(c=|-5-(-4)|=1\). Substitute c into the relationship we found earlier to get a. So, \(a=5*1=5\).

Step 3: Calculate b

The value of b (the semi-minor axis length) could be calculated using the relationship between a, b and c in ellipses: \(a^2 = b^2 + c^2\). Substituting the found values of a and c, we find that \(b^2 = a^2 - c^2 = 5^2 - 1^2 = 24\). Which makes \(b = \sqrt{24}\).

Step 4: Write the Standard Form of the Equation

For a vertically oriented ellipse, the standard form of the equation is \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), where h and k are the coordinates of the center, a is the length of the semi-major axis, and b is the length of the semi-minor axis. Substituting the found values of h,k,a and b, we get: \(\frac{(x-1)^2}{5^2} + \frac{(y+5)^2}{24} = 1\).

Key Concepts

Conic Sections

Conic sections are curves that are formed by intersecting a plane with a double-napped cone. These curves include ellipses, circles, parabolas, and hyperbolas.
Each type of conic section has unique properties and can be defined by specific equations. In the context of an ellipse, a plane intersects a cone in such a way that results in a closed curve. This shape is oval-like and is characterized by its smooth and symmetrical nature.
The ellipse is considered as a subtype of the general conic sections because it retains many core properties of these geometric figures. It maintains a constant sum of distances from any point on the curve to the two focal points, known as foci.

Ellipse Characteristics

Ellipses have distinctive features that differentiate them from other conic sections.
Some of the most critical characteristics of an ellipse include:

• Foci: These are two fixed points on the interior of the ellipse. For our case, the foci are defined to be at (1,-6) and (1,-4). The ellipse in question is vertically oriented because the x-coordinates of its foci are identical.
• Center: The midpoint between the foci, which for this problem is (1,-5).
• Semi-major axis (a): This is the longest radius and is determined to be 5 units about the center. It extends vertically due to the orientation of our ellipse.
• Semi-minor axis (b): The shortest radius, calculated as \(\sqrt{24}\) based on the relationship with \(a\) and \(c\). The equation \(a^2 = b^2 + c^2\) is used to find this.

Every ellipse is symmetrical about both axes and is perfectly balanced around its center.

Standard Form of Ellipse

The standard form of an ellipse's equation provides a simple format to visualize its features more clearly.
There are two variations of the standard form depending on the orientation of the ellipse:

• Horizontal Major Axis: If the ellipse is oriented horizontally, the equation is \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\).
• Vertical Major Axis: For a vertically aligned ellipse, like the one in this problem, the formula changes to \(\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1\).

To write the equation in standard form, start with the center (h,k), which is (1,-5) here, and substitute the values for \(a\) and \(b\).
Using our values of \(a = 5\) and \(b = \sqrt{24}\), we can construct the equation: \(\frac{(x-1)^2}{5^2} + \frac{(y+5)^2}{24} = 1\). This visual representation is crucial for understanding and graphing ellipses.