Problem 34
Question
Find the foci for each equation of an ellipse. $$ 16 x^{2}+4 y^{2}=64 $$
Step-by-Step Solution
Verified Answer
The foci of the ellipse given by the equation \(16x^2 + 4y^2 = 64\) are located at (0, -2\(\sqrt{3}\)) and (0, 2\(\sqrt{3}\)).
1Step 1: Rewrite the Equation in Standard Form
The equation of an ellipse in standard form is given by \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a^2\) and \(b^2\) are the coefficients of \(x^2\) and \(y^2\) respectively. Let's rewrite \(16x^2 + 4y^2 = 64\), dividing each term by 64 to put it in standard form. This gives \(\frac{x^2}{4} + \frac{y^2}{16} = 1\). So, we now have \(a^2 = 16\) and \(b^2 = 4\).
2Step 2: Find the Semi-Major and Semi-Minor Axes
From the standard equation, we get \(a^2 = 16\) and \(b^2 = 4\). Taking the square root of these values gives us \(a = 4\) and \(b = 2\). So, we now have the values of the semi-major axis \(a\) and the semi-minor axis \(b\).
3Step 3: Calculate the Foci
The foci of the ellipse is given by \(\sqrt{a^2 - b^2}\), where \(a\) is the semi-major axis and \(b\) is the semi-minor axis. So, substituting the values of \(a\) and \(b\) into the equation gives \(\sqrt{4^2 - 2^2} = \sqrt{16 - 4} = \sqrt{12}\). Therefore, the foci of the ellipse located at \(0, \pm \sqrt{12}\), having coordinates (0, -2\(\sqrt{3}\)) and (0, 2\(\sqrt{3}\)).
Key Concepts
Equation of an EllipseEllipse FociSemi-Major AxisSemi-Minor Axis
Equation of an Ellipse
The equation of an ellipse is fundamental for understanding its properties.
An ellipse in its standard form is denoted by the equation: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] This representation is key to identifying an ellipse's shape and size. Here, \( a^2 \) and \( b^2 \) are the denominators of the \( x^2 \) and \( y^2 \) terms, respectively. They play a significant role in dictating the length of the ellipse's axes.
An ellipse in its standard form is denoted by the equation: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] This representation is key to identifying an ellipse's shape and size. Here, \( a^2 \) and \( b^2 \) are the denominators of the \( x^2 \) and \( y^2 \) terms, respectively. They play a significant role in dictating the length of the ellipse's axes.
- \( a \): semi-major axis, the longer axis of the ellipse.
- \( b \): semi-minor axis, the shorter axis of the ellipse.
Ellipse Foci
The foci are two special points inside an ellipse. They play a critical role in defining the curve itself.
For any point on the ellipse, the sum of the distances to the foci is constant, and this geometric property is what makes an ellipse distinct from a circle. When you have an ellipse equation in standard form, the foci can be found using the formula \[ \sqrt{a^2 - b^2} \]. Here, \( a \) should always be the semi-major axis, and it's essential to ensure \( a > b \). For the exercise example, where the equation eventually transforms to \[ \frac{x^2}{4} + \frac{y^2}{16} = 1 \], \( a = 4 \) and \( b = 2 \). Thus, the foci are at \[ (0, \pm \sqrt{12}) \] or equivalently \[ (0, \pm 2\sqrt{3}) \].The location of the foci helps to orient the ellipse within a coordinate plane.
For any point on the ellipse, the sum of the distances to the foci is constant, and this geometric property is what makes an ellipse distinct from a circle. When you have an ellipse equation in standard form, the foci can be found using the formula \[ \sqrt{a^2 - b^2} \]. Here, \( a \) should always be the semi-major axis, and it's essential to ensure \( a > b \). For the exercise example, where the equation eventually transforms to \[ \frac{x^2}{4} + \frac{y^2}{16} = 1 \], \( a = 4 \) and \( b = 2 \). Thus, the foci are at \[ (0, \pm \sqrt{12}) \] or equivalently \[ (0, \pm 2\sqrt{3}) \].The location of the foci helps to orient the ellipse within a coordinate plane.
Semi-Major Axis
The semi-major axis is one of the principal axes of an ellipse and is the longest of the two axes.
Its measurement extends from the center of the ellipse to the farthest edge along the major direction. The length of the semi-major axis is represented by \( a \). It is derived by taking the square root of the larger of the two values \( a^2 \) or \( b^2 \) from the standard form.
Its measurement extends from the center of the ellipse to the farthest edge along the major direction. The length of the semi-major axis is represented by \( a \). It is derived by taking the square root of the larger of the two values \( a^2 \) or \( b^2 \) from the standard form.
- In the equation \( \frac{x^2}{4} + \frac{y^2}{16} = 1 \), we found that \( a^2 = 16 \), thus \( a = 4 \).
- This axis lies along the \( y \)-direction in the transformed equation because \( 4^2 = 16 \) is related to the \( y^2 \) term.
Semi-Minor Axis
The semi-minor axis is the shorter axis of the ellipse.
It also stretches from the center to the perimeter of the ellipse, but in the direction of the shortest path. It is designated as \( b \) and establishes the compactness of the ellipse. Specifically, \( b \) is calculated by finding the square root of the smaller value between \( a^2 \) and \( b^2 \).
It also stretches from the center to the perimeter of the ellipse, but in the direction of the shortest path. It is designated as \( b \) and establishes the compactness of the ellipse. Specifically, \( b \) is calculated by finding the square root of the smaller value between \( a^2 \) and \( b^2 \).
- In our specified standard form equation, \( \frac{x^2}{4} + \frac{y^2}{16} = 1 \), we had \( b^2 = 4 \) leading to \( b = 2 \).
- This indicates that the semi-minor axis aligns with the \( x \)-direction.
Other exercises in this chapter
Problem 33
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a. Writing Describe the relationship between the center of a circle and the axes of symmetry of the circle. b. Make a Conjecture Where is the center of an ellip
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