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Difficulty: HardSystems of Linear and Non-Linear Equations

A line with a positive slope passes through the point (0,−5)(0, -5) and is tangent to the parabola y=x2−6x+11y = x^2 - 6x + 11. What is the slope of this line?

Answer: 2

Answer

The slope of the line is 2.
The correct slope is 2. Representing the line as y=mx−5y = mx - 5 and setting it equal to the parabola y=x2−6x+11y = x^2 - 6x + 11 results in the quadratic equation x2−(6+m)x+16=0x^2 - (6 + m)x + 16 = 0. For the line to be tangent, this equation must have exactly one real solution, meaning its discriminant must equal zero: (6+m)2−4(1)(16)=0(6 + m)^2 - 4(1)(16) = 0. Solving this gives 6+m=86 + m = 8 or 6+m=−86 + m = -8, which results in m=2m = 2 or m=−14m = -14. Since the problem specifies that the slope is positive, the value of mm must be 2.

Step-by-Step Solution

1
Write the equation of the line in slope-intercept form.
y=mx−5y = mx - 5, where m>0m > 0.
The line passes through (0,−5)(0, -5), which represents the y-intercept of the line.
2
Equate the line and the parabola to set up the system of equations.
x2−6x+11=mx−5x^2 - 6x + 11 = mx - 5
Setting the two expressions equal allows us to find the x-coordinates of any intersection points.
3
Rearrange the equation into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
x2−(6+m)x+16=0x^2 - (6 + m)x + 16 = 0
Writing the equation in standard form identifies the coefficients a=1a = 1, b=−(6+m)b = -(6 + m), and c=16c = 16 needed for the discriminant.
4
Apply the condition for tangency by setting the discriminant to zero.
b2−4ac=(−(6+m))2−4(1)(16)=0  ⟹  (6+m)2−64=0b^2 - 4ac = (-(6 + m))^2 - 4(1)(16) = 0 \implies (6 + m)^2 - 64 = 0
A line is tangent to a parabola if they touch at exactly one point, meaning the quadratic equation has exactly one real root (discriminant equals zero).
5
Solve the quadratic equation for the slope mm and filter for the positive value.
(6+m)2=64  ⟹  6+m=±8(6+m)^2 = 64 \implies 6+m = \pm 8, yielding m=2m = 2 or m=−14m = -14. Since the slope is positive, m=2m = 2.
Solving the equation gives two possible slope values for tangent lines, and we select the positive slope as specified in the problem statement.

Key Concept

Determining tangency between a linear and quadratic equation by setting the discriminant of the intersection equation to zero.
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Topics

Systems of Linear and Non-Linear Equations

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