Problem 27
Question
In Problems \(23-28,\) use the discriminant to identify the conic without actually graphing. $$ x^{2}+x y+y^{2}-x+2 y+1=0 $$
Step-by-Step Solution
Verified Answer
The conic section is an ellipse.
1Step 1: Understand the Standard Form
The given equation, \(x^2 + xy + y^2 - x + 2y + 1 = 0\), is a second-degree polynomial in two variables \(x\) and \(y\). It's represented as \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\). Identify coefficients: \(A = 1\), \(B = 1\), \(C = 1\). This gives us the formula needed to find the discriminant.
2Step 2: Calculate the Discriminant
The discriminant for a conic section in the form \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\) is given by \(B^2 - 4AC\). Substitute the identified values: \(A = 1\), \(B = 1\), \(C = 1\). Calculate: \(B^2 = 1^2 = 1\) and \(4AC = 4 \cdot 1 \cdot 1 = 4\). Therefore, the discriminant is \(1 - 4 = -3\).
3Step 3: Interpret the Discriminant
The discriminant \(B^2 - 4AC\) helps identify the type of conic section. If \(B^2 - 4AC > 0\), it's a hyperbola; if \(B^2 - 4AC = 0\), it's a parabola; and if \(B^2 - 4AC < 0\), it's an ellipse. Here, \(B^2 - 4AC = -3 < 0\), indicating that the conic is an ellipse.
Key Concepts
Ellipse IdentificationSecond-degree PolynomialsConic Sections
Ellipse Identification
Identifying whether a conic section is an ellipse without graphing can be effectively achieved by examining the discriminant of the equation. The discriminant of a conic section equation is a mathematical expression that can determine the type of conic - such as ellipse, parabola, or hyperbola - based solely on the coefficients of the equation.
The general equation for conic sections is given as: \[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \]
The discriminant, defined as \(B^2 - 4AC\), helps categorize the conic section:
The general equation for conic sections is given as: \[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \]
The discriminant, defined as \(B^2 - 4AC\), helps categorize the conic section:
- If \(B^2 - 4AC > 0\), the conic is a hyperbola.
- If \(B^2 - 4AC = 0\), the conic represents a parabola.
- If \(B^2 - 4AC < 0\), the conic is an ellipse.
Second-degree Polynomials
A second-degree polynomial in two variables, such as the one provided in the exercise, is a fundamental expression in algebra that forms the basis for describing conic sections. A polynomial of this nature has the general form: \[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \]
The polynomials are termed "second-degree" due to the highest power of the variables being two.
Key attributes of second-degree polynomials include:
The polynomials are termed "second-degree" due to the highest power of the variables being two.
Key attributes of second-degree polynomials include:
- They are essential in defining curves and shapes in a plane, through their roots and coefficients.
- The terms \(Ax^2\), \(Bxy\), and \(Cy^2\) with quadratic expressions define the exact type of conic section shape.
- The coefficients \(A\), \(B\), and \(C\) play critical roles in determining the nature of the curve via discriminant analysis.
Conic Sections
Conic sections are shapes created as a plane intersects with the surface of a double cone. These shapes include ellipses, parabolas, and hyperbolas, each with distinctive geometric properties.
The different types of conic sections are:
The different types of conic sections are:
- Ellipses: Formed when the discriminant \(B^2 - 4AC < 0\). They are oval-shaped, representing orbits of planets and shapes with equal distribution.
- Parabolas: Arise when \(B^2 - 4AC = 0\). Parabolas are symmetrical, U-shaped curves that can model projectiles or focus beams of light.
- Hyperbolas: Created when \(B^2 - 4AC > 0\). They consist of two separate curves that mirror each other and can appear in scenarios involving open orbits between celestial bodies.
Other exercises in this chapter
Problem 27
In Problems \(21-44,\) find an equation of the hyperbola that satisfies the given conditions. Center \((0,0),\) one vertex \(\left(0, \frac{5}{2}\right)\) one f
View solution Problem 27
Find an equation of the ellipse that satisfies the given conditions. Vertices \((-3,-3),(5,-3),\) endpoints of minor axis (1,-1),(1,-5)
View solution Problem 27
Find an equation of parabola that satisfies the given conditions. Focus \((-4,0), \operatorname{directrix} x=4\)
View solution Problem 28
Find an equation of the ellipse that satisfies the given conditions. Vertices \((1,-6),(1,2),\) endpoints of minor axis (-2,-2),(4,-2)
View solution