Problem 56 - College Algebra

Question

For the following exercises, determine the value of \(k\) based on the given equation. Given \(4 x^{2}+k x y+16 y^{2}+8 x+24 y-48=0\) find \(k\) for the graph to be a parabola.

Step-by-Step Solution

Answer

\( k = \pm 16 \)

Step 1: Identifying the Conic Section

The given equation is in the form of a general second degree equation: \[ 4x^2 + kxy + 16y^2 + 8x + 24y - 48 = 0 \]In order for this equation to represent a parabola, the discriminant \( B^2 - 4AC \) must equal zero, where \( A \), \( B \), and \( C \) are coefficients of the terms \( x^2 \), \( xy \), and \( y^2 \), respectively.

Step 2: Assigning Coefficients

In the equation, the coefficients are:\( A = 4 \), \( B = k \), and \( C = 16 \). The discriminant condition for a parabola is:\[ B^2 - 4AC = 0 \]

Step 3: Setting Up the Equation

Substitute the values of \( A \), \( B \), and \( C \) into the discriminant:\[ k^2 - 4(4)(16) = 0 \]

Step 4: Solving for 'k'

Simplified, the equation becomes:\[ k^2 - 256 = 0 \]Add 256 to both sides:\[ k^2 = 256 \]Take the square root of both sides to solve for \( k \):\[ k = \pm 16 \]

Key Concepts

Parabola

A parabola is a unique and simple curve that has special properties when it comes to conic sections. It can be visualized as the path traced by a projectile, a satellite dish, or even the arches of some bridges. In mathematics, a parabola is a U-shaped curve on a plane.
A conic section can form different shapes: ellipse, hyperbola, or parabola, depending on its discriminant. A parabola specifically is produced when the equation’s discriminant equals zero.
For an equation of the standard form:

• General form: \[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \]
• To represent a parabola, the discriminant \[ B^2 - 4AC \] must be zero.

Think about a quadratic equation; its graph, a special case, is a parabola. For our given equation, when solved for the proper value of a coefficient, it sticks to this rule, ensuring it forms a parabola.

Discriminant

The discriminant is a crucial tool in identifying the type of conic section represented by a second-degree equation. Let's dig deeper into how it works!
The discriminant for conic equations like \[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \] is given by:

• \[ D = B^2 - 4AC \]
• If \[ D = 0 \], you have a parabola.
• If \[ D > 0 \], it's a hyperbola.
• If \[ D < 0 \], it's an ellipse.

For our equation \[ 4x^2 + kxy + 16y^2 + 8x + 24y - 48 = 0 \], the discriminant depends on the coefficients, specifically, ensuring \[ B^2 - 4AC = 0 \]. Here, manipulating \[ B = k \] with the given coefficients aims to make \[ B^2 - 4(4)(16) = 0 \], which then gives us the solutions for \[ k \]. Understanding these relationships helps determine what kind of curve the equation describes.

Coefficients

Coefficients in a conic equation control the properties and type of the curve it forms, just like the knobs of a music equalizer shape the sound. Different coefficients can tweak a second degree polynomial to represent parabolas, ellipses, or hyperbolas.
For our specific equation \[ 4x^2 + kxy + 16y^2 + 8x + 24y - 48 = 0 \], the key coefficients to determine the curve type are:

• \[ A = 4 \] for \[ x^2 \]
• \[ B = k \] for \[ xy \]
• \[ C = 16 \] for \[ y^2 \]

These coefficients define how the conic curve behaves. To make sure the equation forms a parabola, the relationship between these coefficients is subjected to the discriminant condition. Solving\[ k^2 - 256 = 0 \] determines \[ k = \pm 16 \], showing how the coefficient \[ k \] influences the equation to align with the condition for a parabola.