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Difficulty: Very hardSystems of Linear and Non-Linear Equations

A small drone's path in a vertical plane is modeled by the equation y=3x2−4x+2y = 3x^2 - 4x + 2, where xx is the horizontal distance in meters and yy is the height in meters. A laser beam travels along a straight line in the same plane such that the sum of twice its horizontal distance and its height is a constant cc, where both are in meters. The laser beam intersects the drone's path at two distinct points. If the distance between these two intersection points is 523\frac{5\sqrt{2}}{3} meters, what is the value of 2c2c?

  1. A
    66
  2. B
    1010
  3. 55Answer
  4. D
    −13-\frac{1}{3}
  5. E
    53\frac{5}{3}

Answer

The correct value of 2c2c is 55.
The correct value of 2c2c is 55. Substituting the linear equation y=c−2xy = c - 2x into the quadratic equation y=3x2−4x+2y = 3x^2 - 4x + 2 yields 3x2−2x+(2−c)=03x^2 - 2x + (2-c) = 0. The distance between the intersection points is d=∣x2−x1∣5=523d = |x_2 - x_1|\sqrt{5} = \frac{5\sqrt{2}}{3}, which simplifies to (x2−x1)2=109(x_2 - x_1)^2 = \frac{10}{9}. Using the identity (x2−x1)2=(x1+x2)2−4x1x2(x_2 - x_1)^2 = (x_1 + x_2)^2 - 4x_1 x_2 and Vieta's formulas, we find the equation 49−8−4c3=109\frac{4}{9} - \frac{8-4c}{3} = \frac{10}{9}. Solving this equation yields c=52c = \frac{5}{2}, and thus 2c=52c = 5.

Step-by-Step Solution

1
Set up the system of equations.
The drone's path is y=3x2−4x+2y = 3x^2 - 4x + 2 and the laser's path is 2x+y=c  ⟹  y=−2x+c2x + y = c \implies y = -2x + c.
This represents the mathematical formulation of both paths in the vertical plane.
2
Equate the equations to find the x-coordinates of the intersection points.
3x2−4x+2=−2x+c  ⟹  3x2−2x+(2−c)=03x^2 - 4x + 2 = -2x + c \implies 3x^2 - 2x + (2-c) = 0.
The intersection points satisfy both equations, so we can solve for xx by substitution.
3
Express the distance between the intersection points P(x1,y1)P(x_1, y_1) and Q(x2,y2)Q(x_2, y_2) using the slope.
d=∣x2−x1∣5d = |x_2 - x_1|\sqrt{5}.
Since the points lie on the line with slope −2-2, we have y2−y1=−2(x2−x1)y_2 - y_1 = -2(x_2 - x_1). The distance formula becomes d=(x2−x1)2+(−2(x2−x1))2=5(x2−x1)2=∣x2−x1∣5d = \sqrt{(x_2 - x_1)^2 + (-2(x_2 - x_1))^2} = \sqrt{5(x_2 - x_1)^2} = |x_2 - x_1|\sqrt{5}.
4
Equate the distance expression to the given distance to find (x2−x1)2(x_2 - x_1)^2.
(x2−x1)2=109(x_2 - x_1)^2 = \frac{10}{9}.
We are given d=523d = \frac{5\sqrt{2}}{3}. Setting ∣x2−x1∣5=523|x_2 - x_1|\sqrt{5} = \frac{5\sqrt{2}}{3} and squaring both sides gives 5(x2−x1)2=5095(x_2 - x_1)^2 = \frac{50}{9}, which simplifies to (x2−x1)2=109(x_2 - x_1)^2 = \frac{10}{9}.
5
Apply Vieta's formulas and the algebraic identity for (x2−x1)2(x_2 - x_1)^2.
109=49−8−4c3\frac{10}{9} = \frac{4}{9} - \frac{8-4c}{3}.
For the quadratic equation 3x2−2x+(2−c)=03x^2 - 2x + (2-c) = 0, we have x1+x2=23x_1 + x_2 = \frac{2}{3} and x1x2=2−c3x_1 x_2 = \frac{2-c}{3}. We use the identity (x2−x1)2=(x1+x2)2−4x1x2(x_2 - x_1)^2 = (x_1 + x_2)^2 - 4x_1 x_2.
6
Solve for cc and find 2c2c.
c=52  ⟹  2c=5c = \frac{5}{2} \implies 2c = 5.
Multiplying the equation by 99 gives 10=4−3(8−4c)  ⟹  10=−20+12c  ⟹  12c=30  ⟹  c=5210 = 4 - 3(8-4c) \implies 10 = -20 + 12c \implies 12c = 30 \implies c = \frac{5}{2}. Therefore, 2c=52c = 5.

Key Concept

Solving systems of linear and non-linear equations using substitution, coordinate geometry distance formula, and quadratic root relationships.
Estimated Time:3m 0s
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Topics

Systems of Linear and Non-Linear Equations

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