Problem 26
Question
Exer. 19-30: Find an equation for the ellipse that has its center at the origin and satisfies the given conditions. Eccentricity \(\frac{1}{2}\), vertices on the \(x\)-axis, passing through \((1,3)\)
Step-by-Step Solution
Verified Answer
The equation of the ellipse is \(\frac{x^2}{4} + \frac{y^2}{3} = 1\).
1Step 1: Understanding the Problem
The ellipse is centered at the origin and has a given eccentricity of \(\frac{1}{2}\). Its vertices are along the \(x\)-axis, which tells us that the major axis is horizontal. We are given a point \((1,3)\) on the ellipse.
2Step 2: Applying Eccentricity Formula
For an ellipse with center at the origin and major axis along \(x\)-axis, the equation is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(c^2 = a^2 - b^2\). Here, the eccentricity \(e\) is \(\frac{c}{a}\). With \(e = \frac{1}{2}\), we have \(c = \frac{a}{2}\).
3Step 3: Substituting Eccentricity
Substituting \(c = \frac{a}{2}\) into \(c^2 = a^2 - b^2\), gives \(\left(\frac{a}{2}\right)^2 = a^2 - b^2\). Simplify to get \(\frac{a^2}{4} = a^2 - b^2\), or \(b^2 = \frac{3}{4}a^2\).
4Step 4: Using Given Point
Substitute \(x = 1\) and \(y = 3\) into the ellipse equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). This gives \(\frac{1}{a^2} + \frac{9}{b^2} = 1\). Substitute \(b^2 = \frac{3}{4}a^2\) into the equation.
5Step 5: Solving for \(a^2\)
From \(\frac{1}{a^2} + \frac{9}{\frac{3}{4}a^2} = 1\), simplify to \(\frac{1}{a^2} + \frac{12}{4a^2} = 1\) or \(\frac{1}{a^2} + \frac{3}{a^2} = 1\). This results in \(\frac{4}{a^2} = 1\), hence \(a^2 = 4\).
6Step 6: Finding \(b^2\)
Using \(b^2 = \frac{3}{4}a^2\) and \(a^2 = 4\), substitute to get \(b^2 = \frac{3}{4} \times 4 = 3\).
7Step 7: Conclusion
Now we have \(a^2 = 4\) and \(b^2 = 3\). The equation of the ellipse is \(\frac{x^2}{4} + \frac{y^2}{3} = 1\).
Key Concepts
EccentricityEllipse Centered at OriginVertices on x-axis
Eccentricity
Eccentricity is a measure that describes how stretched out an ellipse appears. It tells us about the shape of the ellipse. In mathematical terms, eccentricity (\(e\)) is the ratio of the distance from the center of the ellipse to its focus (\(c\)) compared to the distance from the center to a vertex on the major axis (\(a\)).
For ellipses:
For ellipses:
- The eccentricity (\(e\)) is always between 0 and 1, where 0 is a perfect circle.
- In this exercise, given eccentricity is \(\frac{1}{2}\).
- We use the formula \(e = \frac{c}{a}\) to find the relationship between these distances.
Ellipse Centered at Origin
Placing an ellipse centered at the origin simplifies its equation. The center of an ellipse is the midpoint of both its major and minor axes.
At the origin, the general equation of an ellipse is:\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]Here:
At the origin, the general equation of an ellipse is:\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]Here:
- \(a\) refers to the semi-major axis length.
- \(b\) is the semi-minor axis length.
Vertices on x-axis
When the vertices of an ellipse lie on the x-axis, it implies that the major axis runs horizontally. This orientation affects the standard form of the ellipse equation because:
Ultimately, when working out any point through such an ellipse, this information reflects the breadth and orientation essential in forming its graph and understanding its physical arrangement in space.
- The major axis is tied with \(x\), meaning \(a\) belongs in the \(x^2\) denominator.
- The vertex positions directly affect values of \(a\) since they are located at \((-a, 0)\) and \((a, 0)\).
Ultimately, when working out any point through such an ellipse, this information reflects the breadth and orientation essential in forming its graph and understanding its physical arrangement in space.
Other exercises in this chapter
Problem 26
Find an equation for the hyperbola that has its center at the origin and satisfies the given conditions. Vertices \(V(\pm 4,0)\), passing through \((8,2)\)
View solution Problem 26
Exer. 19-30: Find an equation of the parabola that satisfies the given conditions. Vertex \(V(1,-2)\), focus \(F(1,0)\)
View solution Problem 27
Exer. 25-32: Find a polar equation of the conic with focus at the pole that has the given eccentricity and equation of directrix. $$ e=\frac{4}{3}, \quad r \cos
View solution Problem 27
Exer. 27-44: Find an equation in \(x\) and \(y\) that has the same graph as the polar equation. Use it to help sketch the graph in an \(r \theta\)-plane. $$ r \
View solution