Problem 57 - Algebra 2

Question

Find the vertices and co-vertices of each ellipse. $$ 25 x^{2}+16 y^{2}=1600 $$

Step-by-Step Solution

Answer

The vertices of the ellipse are at points (8, 0) and (-8, 0), and the co-vertices are at points (0, 10) and (0, -10).

Step 1: Identify and calculate the values of a and b

The general formula for an ellipse is \( \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \), so comparing this with the given equation, we equate the denominators of \( x^{2} \) and \( y^{2} \) respectively to 25 and 16, to get the square of axes lengths. We therefore have \( a^{2} = \frac{1600}{25} = 64 \) and \( b^{2} = \frac{1600}{16} = 100 \). Taking the square root of each, we find \( a = \sqrt{64} = 8 \) and \( b = \sqrt{100} = 10 \).

Step 2: Determine the vertices

The vertices of the ellipse are determined using the semi-major axis length a. Substituting \( a = 8 \) into the vertices formula \( (±a, 0) \), we get the vertices at points (8, 0) and (-8, 0).

Step 3: Determine the co-vertices

The co-vertices of the ellipse are determined using the semi-minor axis length b. Substituting \( b = 10 \) into the co-vertices formula \( (0, ±b) \), we get the co-vertices at points (0, 10) and (0, -10).

Key Concepts

Vertices of an Ellipse

To find the vertices of an ellipse, it's essential to understand what they are and how they relate to the ellipse's shape. An ellipse is similar to a stretched circle and is defined by its two fixed points called foci.
The longest distance across an ellipse, passing through its center, is called the major axis. The endpoints of this axis are the vertices.

• To determine the vertices from an equation, we first need the ellipse in its standard form: \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]
• If our given ellipse equation is in another form, such as \[25x^2 + 16y^2 = 1600,\]we will need to rearrange it by dividing every term by 1600 to get it in the standard format.
• This step allows us to identify the semi-major axis length, \(a\), from the term with the larger denominator.
• Once \(a\) is found, the vertices can be calculated as \((\pm a, 0)\) for an ellipse aligned with the x-axis, giving us the points where the ellipse stretches the farthest horizontally.

Co-vertices of an Ellipse

The co-vertices of an ellipse are located on the minor axis. This axis is the shortest path across the ellipse through its center.
The semi-minor axis length, \(b\), determines where these co-vertices are positioned.

• In the standard ellipse equation, \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,\]the term with the smaller denominator corresponds to the semi-minor axis.
• For the given example, \[25x^2 + 16y^2 = 1600,\] the standard form reveals \(b^2\) by rearranging as \[\frac{x^2}{64} + \frac{y^2}{100} = 1.\]
• Thus, the semi-minor axis \(b\) is determined by \(\sqrt{100} = 10\).
• We use this to find the co-vertices at locations \((0, \pm b)\) when aligned to the y-axis.
• These co-vertices represent the height where the ellipse is the narrowest, but still stretches vertically.

Equation of an Ellipse

Understanding the equation of an ellipse helps see the shape's properties and dimensions. The "standard form" equation provides all necessary information to describe the ellipse's size and orientation.

• The standard equation for an ellipse centered at the origin is:\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,\]where \(a\) is the semi-major axis length and \(b\) is the semi-minor axis length.
• Depending on which of \(a^2\) or \(b^2\) is larger, the ellipse can be "stretched" either horizontally or vertically.
• For our given equation \[25x^2 + 16y^2 = 1600,\]it requires reorganization into the standard form by dividing each term by 1600:\[\frac{x^2}{64} + \frac{y^2}{100} = 1.\]
• This tells us everything about the ellipse's axes, where \(a = 8\) and \(b = 10\), showing it's oriented vertically since \(b > a\) here.
• Understanding this form helps easily visualize and plot the ellipse, identifying its vertices and co-vertices as key points describing its overall dimensions.