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Difficulty: MediumNonlinear Systems of Equations

A parabola and a line intersect at exactly one point in the xyxy-plane. The parabola is defined by the equation y=−x2+6x−2y = -x^2 + 6x - 2 and the line is defined by the equation y=2x+ky = 2x + k, where kk is a constant. What is the value of kk?

Answer: 2

Answer

The correct answer is 2.
To find the value of kk where the parabola and the line intersect at exactly one point, we equate the two equations: −x2+6x−2=2x+k-x^2 + 6x - 2 = 2x + k. Rearranging this into standard quadratic form yields x2−4x+(k+2)=0x^2 - 4x + (k + 2) = 0. For a quadratic equation to have exactly one real solution, its discriminant, b2−4acb^2 - 4ac, must be equal to zero. Substituting a=1a = 1, b=−4b = -4, and c=k+2c = k + 2 into the discriminant formula gives (−4)2−4(1)(k+2)=0(-4)^2 - 4(1)(k + 2) = 0, which simplifies to 16−4k−8=016 - 4k - 8 = 0, or 8−4k=08 - 4k = 0. Solving for kk gives k=2k = 2.

Step-by-Step Solution

1
Equate the equations of the parabola and the line to set up an equation for the x-coordinates of their intersection points.
−x2+6x−2=2x+k-x^2 + 6x - 2 = 2x + k
At the points of intersection, the y-values of both equations are equal.
2
Rearrange the equation into the standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
x2−4x+(k+2)=0x^2 - 4x + (k + 2) = 0
This allows us to identify the coefficients a=1a = 1, b=−4b = -4, and c=k+2c = k + 2 to apply the quadratic discriminant.
3
Set the discriminant b2−4acb^2 - 4ac equal to zero.
(−4)2−4(1)(k+2)=0(-4)^2 - 4(1)(k + 2) = 0
A quadratic system has exactly one real solution (tangency) if and only if the discriminant of the resulting quadratic equation is zero.
4
Solve the linear equation for kk.
k=2k = 2
Simplifying the expression yields 16−4k−8=016 - 4k - 8 = 0, which simplifies to 8−4k=08 - 4k = 0, giving k=2k = 2.

Key Concept

Nonlinear Systems of Equations
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Topics

Nonlinear Systems of Equations

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