Problem 44
Question
Find an equation for the ellipse that satisfies the given conditions. Eccentricity: \(\sqrt{3} / 2,\) foci on \(y\) -axis, length of major axis: 4
Step-by-Step Solution
Verified Answer
The equation is \( x^2 + \frac{y^2}{4} = 1 \).
1Step 1: Identify the ellipse orientation
Since the foci are on the y-axis, the major axis is vertical. Thus, the standard form of the ellipse is \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \) where \( a > b \).
2Step 2: Use major axis to find 'a'
The length of the major axis is given as 4. Thus, the semi-major axis \( a \) is half of that: \( a = 2 \).
3Step 3: Relate eccentricity to find 'c'
The eccentricity \( e \) is given by \( e = \frac{c}{a} \). We are given \( e = \frac{\sqrt{3}}{2} \). So, \( \frac{c}{2} = \frac{\sqrt{3}}{2} \), which gives \( c = \sqrt{3} \).
4Step 4: Use the relationship between a, b, and c
For an ellipse, \( c^2 = a^2 - b^2 \). Substituting \( c = \sqrt{3} \) and \( a = 2 \), we have \( (\sqrt{3})^2 = 2^2 - b^2 \). Therefore, \( 3 = 4 - b^2 \). Solving for \( b^2 \), we find \( b^2 = 1 \).
5Step 5: Write the equation of the ellipse
Now that we have \( a^2 = 4 \) and \( b^2 = 1 \), the equation of the ellipse is: \( \frac{x^2}{1} + \frac{y^2}{4} = 1 \), simplifying to \( x^2 + \frac{y^2}{4} = 1 \).
Key Concepts
EccentricityFociMajor AxisSemi-Major Axis
Eccentricity
Eccentricity is a fundamental concept when it comes to ellipses, describing how much the ellipse is stretched out compared to a perfect circle. An ellipse with an eccentricity of zero is a perfect circle, while one with an eccentricity close to one is more elongated. The eccentricity, denoted as \( e \), defines the shape of the ellipse and is calculated using the formula:
- \( e = \frac{c}{a} \)
Foci
The foci (plural of focus) are two specific points located inside every ellipse. These points play a crucial role in the definition and properties of the ellipse. An ellipse can be defined as the set of all points such that the sum of the distances from two fixed points (the foci) is constant.In a standard vertical ellipse like in our example, the foci lie on the y-axis. The distance \( c \) from the center of the ellipse to each focus is a measure used to position them. Thus, if the center of the ellipse is at the origin (0,0), the coordinates for the foci are:
- (0, \( c \))
- (0, -\( c \))
Major Axis
The major axis of an ellipse is its longest diameter, passing through the center and both foci. This axis determines the overall length and orientation of the ellipse.For ellipses where the foci are on the y-axis, the major axis is vertical. The length of the major axis is given directly, but it is twice the length of the semi-major axis \( a \), so:
- Length of major axis = \( 2a \)
Semi-Major Axis
The semi-major axis is half of the major axis, thus representing the largest radius of the ellipse. It is denoted by \( a \) and is critical in defining the ellipse's dimensions and position.The semi-major axis is used in the primary equation of an ellipse:
- \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \)
- Helps calculate eccentricity \( e \)
- Links with \( b \) for complete equation of the ellipse
Other exercises in this chapter
Problem 43
Find an equation for the ellipse that satisfies the given conditions. Eccentricity: \(0.8,\) foci: \(( \pm 1.5,0)\)
View solution Problem 44
The hyperbolas $$ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \quad \text { and } \quad \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=-1 $$ are said to be conjugate to
View solution Problem 45
In the derivation of the equation of the hyperbola at the beginning of this section, we said that the equation $$ \sqrt{(x+c)^{2}+y^{2}}-\sqrt{(x-c)^{2}+y^{2}}=
View solution Problem 45
Find the intersection points of the pair of ellipses. Sketch the graphs of each pair of equations on the same coordinate axes, and label the points of intersect
View solution