Problem 17
Question
Identifying \(A=B=C=1,\) we compute \(B^{2}-4 A C=-3 < 0 .\) The equation is elliptic.
Step-by-Step Solution
Verified Answer
The equation represents an ellipse because \( B^2 - 4AC < 0 \).
1Step 1: Identify Given Values
We are given that \( A = B = C = 1 \). This means we have the coefficients for identifying the type of conic section.
2Step 2: Compute \( B^2 - 4AC \)
The formula \( B^2 - 4AC \) helps determine the type of conic. For our values, this computes to \( 1^2 - 4 \times 1 \times 1 = 1 - 4 = -3 \).
3Step 3: Determine the Significance of \( B^2 - 4AC \)
The value \( B^2 - 4AC = -3 \) is less than 0, which indicates that the conic section is an ellipse.
Key Concepts
Conic SectionsEllipsesAlgebraic Equations
Conic Sections
Conic sections are curves obtained by intersecting a plane with a cone. Depending on the angle at which the plane intersects the cone, we can get different shapes: circles, ellipses, parabolas, or hyperbolas. These shapes are significant in various mathematical and physical applications.
To classify a conic section, mathematicians use the general quadratic equation: \[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \]The type of conic section is determined by the discriminant of the conic, which is defined by the expression \( B^2 - 4AC \). Each conic section has distinct features based on the value:
To classify a conic section, mathematicians use the general quadratic equation: \[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \]The type of conic section is determined by the discriminant of the conic, which is defined by the expression \( B^2 - 4AC \). Each conic section has distinct features based on the value:
- If \( B^2 - 4AC > 0 \), the conic is a hyperbola.
- If \( B^2 - 4AC = 0 \), the conic is a parabola.
- If \( B^2 - 4AC < 0 \), the conic is an ellipse, as in our given exercise.
Ellipses
Ellipses are one form of conic sections characterized by the discriminant \( B^2 - 4AC < 0 \), as in our exercise. An ellipse is a set of points on a plane, such that the sum of the distances of each point from two fixed points (foci) is constant.
The standard equation of an ellipse centered at the origin \((h, k)\) takes the form: \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]where \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively.
Some important properties of ellipses include:
The standard equation of an ellipse centered at the origin \((h, k)\) takes the form: \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]where \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively.
Some important properties of ellipses include:
- The longest diameter is called the major axis.
- The shortest diameter is called the minor axis.
- The points where the ellipse intersects its axes are called the vertices.
Algebraic Equations
Algebraic equations relate variables and constants using algebraic operations (addition, subtraction, multiplication, and division). In the context of conic sections, algebraic equations represent the curves on a graph.
The general form of an algebraic equation for conics is:\[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \]Here, each term represents an element of the ellipse, circle, parabola, or hyperbola, depending on the discriminant \( B^2 - 4AC \).
To determine the type of conic section from the equation, evaluate the coefficients \( A, B, \) and \( C \) using the discriminant
The general form of an algebraic equation for conics is:\[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \]Here, each term represents an element of the ellipse, circle, parabola, or hyperbola, depending on the discriminant \( B^2 - 4AC \).
To determine the type of conic section from the equation, evaluate the coefficients \( A, B, \) and \( C \) using the discriminant
- For an ellipse, the key condition is \( B^2 - 4AC < 0 \).
- For specifically horizontal or vertical ellipses, \( B = 0 \).
Other exercises in this chapter
Problem 16
Referring to the discussion in this section of the text we identify \(a=b=2, f(x)=0,\) $$g(x)=\left\\{\begin{array}{ll} x, & 0
View solution Problem 16
Substituting \(u(x, t)=X(x) T(t)\) into the partial differential equation yields \(a^{2} X^{\prime \prime} T-g=X T^{\prime \prime},\) which is not separable.
View solution Problem 18
Identifying \(A=3, B=5,\) and \(C=1,\) we compute \(B^{2}-4 A C=13 > 0 .\) The equation is hyperbolic.
View solution Problem 19
Identifying \(A=1, B=6,\) and \(C=9,\) we compute \(B^{2}-4 A C=0 .\) The equation is parabolic.
View solution