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Zorluk: Çok zorSystems of Linear and Non-Linear Equations

A line with a positive slope passes through the point (0,−4)(0, -4) and is tangent to the circle x2+y2=4x^2 + y^2 = 4. If this same line is also tangent to the parabola y=x2+ky = x^2 + k, what is the value of the constant kk?

Cevap: -3.25

Cevap

The constant kk must be −3.25-3.25.
By writing the equation of the line passing through (0,−4)(0, -4) as y=mx−4y = mx - 4 and applying the condition that it is tangent to the circle x2+y2=4x^2 + y^2 = 4, we find the positive slope is m=3m = \sqrt{3}. Substituting this tangent line y=3x−4y = \sqrt{3}x - 4 into the parabola equation y=x2+ky = x^2 + k yields the quadratic equation x2−3x+(k+4)=0x^2 - \sqrt{3}x + (k + 4) = 0. For the line to be tangent to the parabola, the discriminant of this equation must be zero, which gives 3−4(k+4)=03 - 4(k + 4) = 0, leading to the final value k=−3.25k = -3.25.

Adım Adım Çözüm

1
Set up the equation of the line passing through (0,−4)(0, -4) and substitute it into the equation of the circle.
x2+(mx−4)2=4  ⟹  (1+m2)x2−8mx+12=0x^2 + (mx - 4)^2 = 4 \implies (1 + m^2)x^2 - 8mx + 12 = 0
This establishes the quadratic equation representing the intersection points of the line and the circle.
2
Solve for the slope mm by setting the discriminant of the intersection equation to zero.
(−8m)2−4(1+m2)(12)=0  ⟹  64m2−48−48m2=0  ⟹  16m2=48  ⟹  m=3(-8m)^2 - 4(1 + m^2)(12) = 0 \implies 64m^2 - 48 - 48m^2 = 0 \implies 16m^2 = 48 \implies m = \sqrt{3}
Since the line is tangent to the circle, the intersection equation must have exactly one real solution, meaning its discriminant is zero. The positive root is selected because the slope is specified as positive.
3
Equate the line y=3x−4y = \sqrt{3}x - 4 and the parabola y=x2+ky = x^2 + k to analyze their intersection.
x2−3x+(k+4)=0x^2 - \sqrt{3}x + (k + 4) = 0
This sets up the quadratic equation for the intersection points between the tangent line and the parabola.
4
Solve for kk by setting the discriminant of this second quadratic equation to zero.
(−3)2−4(1)(k+4)=0  ⟹  3−4k−16=0  ⟹  −13−4k=0  ⟹  k=−3.25(-\sqrt{3})^2 - 4(1)(k + 4) = 0 \implies 3 - 4k - 16 = 0 \implies -13 - 4k = 0 \implies k = -3.25
For the line to be tangent to the parabola, the system must have exactly one solution, meaning the discriminant of their intersection equation must be zero.

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