Problem 7
Question
Identify each equation as an ellipse or a hyperbola. $$\frac{x^{2}}{25}+y^{2}=1$$
Step-by-Step Solution
Verified Answer
The given equation \(\frac{x^2}{25} + y^2 = 1\) is an ellipse, as both x and y terms are positive and it matches the general form of an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).
1Step 1: Identify the general form of the given equation
We are given the equation \(\frac{x^2}{25} + y^2 = 1\).
2Step 2: Analyze signs of the terms
Look for the signs of the terms involving x and y. We have the equation \(\frac{x^2}{25} + y^2 = 1\).
Both terms are positive as there is no negative sign before any of them.
3Step 3: Compare the equation with the general forms of ellipse and hyperbola equations
The given equation has the form \(\frac{x^2}{a^2} + y^2 = 1\) which is the general form of an ellipse.
4Step 4: Conclusion
The given equation \(\frac{x^2}{25} + y^2 = 1\) is an ellipse.
Key Concepts
EllipseHyperbolaEquation AnalysisGeneral Form of Conic Sections
Ellipse
An ellipse is a special type of conic section that looks like a stretched circle. Ellipses have several key properties that define their shape and dimensions. The defining feature of an ellipse is that the sum of the distances from any point on the ellipse to two fixed points, called foci, is constant. Ellipses can be oriented either horizontally or vertically.
- The standard form of an ellipse is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) if centered at the origin.
- Here, \(a\) and \(b\) are the lengths from the center to the ellipse along the x-axis and y-axis respectively.
- If \(a > b\), the ellipse is elongated along the x-axis. Conversely, if \(b > a\), it is elongated along the y-axis.
For example, the equation from the problem, \(\frac{x^2}{25} + y^2 = 1\), can be rewritten as \(\frac{x^2}{5^2} + \frac{y^2}{1^2} = 1\), indicating an ellipse stretched along the x-axis.
Hyperbola
A hyperbola is another type of conic section that appears as two separate curves or "arms". Unlike ellipses, hyperbolas have different properties and equations.
- In standard form, a hyperbola can be written as \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) or \(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\).
- The key distinction of hyperbolas is the presence of a subtraction between the terms, indicating the hyperbola's separate branches.
- Hyperbolas have two foci and a difference in distances from any point on the hyperbola to these foci, which is constant.
Equation Analysis
When analyzing conic section equations, identifying the signs and coefficients in the equation is crucial. The key aspect to look for is which conic form the equation resembles.
- If all the terms are positive, and it follows the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), then the conic is an ellipse.
- If there is a subtraction, \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) or \(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\), it represents a hyperbola.
General Form of Conic Sections
The general equation for conic sections provides a unified way of expressing ellipses, hyperbolas, circles, and parabolas. This form is versatile but identifying each conic section requires attention to specific characteristics within the equation.
- The general quadratic form is given by \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\).
- Different values and combinations of \(A\), \(B\), and \(C\), as well as the absence or presence of certain terms, determine the type of conic.
- If \(B = 0\) and \(A, C > 0\) with the equation being positive, often it describes an ellipse.
- If \(B = 0\) and \(A, C\) have opposite signs, it represents a hyperbola.
Other exercises in this chapter
Problem 7
Solve each quadratic inequality. Graph the solution set and write the solution in interval notation. $$x^{2}+6 x-7 \geq 0$$
View solution Problem 7
Solve each system. $$\begin{aligned} x^{2}+4 y &=8 \\ x+2 y &=-8 \end{aligned}$$
View solution Problem 7
Identify the center and radius of each circle and graph. $$(x+3)^{2}+y^{2}=4$$
View solution Problem 8
Solve the system of equations. $$\begin{aligned} x+2 y-z &=4 \\ 4 x-3 y+z &=8 \\ 5 x-y &=12 \end{aligned}$$
View solution