Tangency and Quadratic Curves

Oct 13, 2025 by windailys.com 125 views

Introduction

In this article, we will explore the problem of finding the values of k for which the x-axis is a tangent to the curve f(x) = kx^2 + (1+k)x + k. We will also determine the equations of the curve for these values of k. To do this, we will use the concept of tangency and the properties of quadratic curves.

Tangency and Quadratic Curves

A tangent to a curve is a line that touches the curve at a single point. In the case of a quadratic curve, the tangent line will have a slope equal to the derivative of the curve at the point of tangency. For the curve f(x) = kx^2 + (1+k)x + k, the derivative is given by f'(x) = 2kx + (1+k).

Condition for Tangency

For the x-axis to be a tangent to the curve, the curve must intersect the x-axis at a single point. This means that the equation f(x) = 0 must have a single solution. To find this solution, we set f(x) = 0 and solve for x:

kx^2 + (1+k)x + k = 0

Using the quadratic formula, we get:

x = (-b ± √(b^2 - 4ac)) / 2a

where a = k, b = (1+k), and c = k. For the x-axis to be a tangent, the discriminant (b^2 - 4ac) must be equal to zero:

(1+k)^2 - 4k^2 = 0

Expanding and simplifying, we get:

1 + 2k + k^2 - 4k^2 = 0

Combine like terms:

-3k^2 + 2k + 1 = 0

Solving the Quadratic Equation

To solve the quadratic equation -3k^2 + 2k + 1 = 0, we can use the quadratic formula:

k = (-b ± √(b^2 - 4ac)) / 2a

where a = -3, b = 2, and c = 1. Plugging in these values, we get:

k = (-(2) ± √((2)^2 - 4(-3)(1))) / (2(-3))

Simplify:

k = (-2 ± √(4 + 12)) / (-6)

k = (-2 ± √16) / (-6)

k = (-2 ± 4) / (-6)

Solving for k

We have two possible values for k:

k = (-2 + 4) / (-6) or k = (-2 - 4) / (-6)

Simplify:

k = 2 / (-6) or k = -6 / (-6)

k = -1/3 or k = 1

Equations of the Curve

Now that we have found the values of k, we can determine the equations of the curve. For k = -1/3, we have:

f(x) = (-1/3)x^2 + (1-1/3)x + (-1/3)

Simplify:

f(x) = (-1/3)x^2 + (2/3)x - 1/3

For k = 1, we have:

f(x) = x^2 + (1+1)x + 1

Simplify:

f(x) = x^2 + 2x + 1

Conclusion

In this article, we have found the values of k for which the x-axis is a tangent to the curve f(x) = kx^2 + (1+k)x + k. We have also determined the equations of the curve for these values of k. The values of k are -1/3 and 1, and the equations of the curve are f(x) = (-1/3)x^2 + (2/3)x - 1/3 and f(x) = x^2 + 2x + 1, respectively.

Discussion

The problem of finding the values of k for which the x-axis is a tangent to the curve f(x) = kx^2 + (1+k)x + k is an interesting one. It requires the use of the concept of tangency and the properties of quadratic curves. The solution involves solving a quadratic equation and determining the equations of the curve for the values of k.

References

[1] "Quadratic Curves" by Math Open Reference
[2] "Tangency" by Wolfram MathWorld

Note

Q: What is the condition for the x-axis to be a tangent to the curve f(x) = kx^2 + (1+k)x + k?

A: The x-axis will be a tangent to the curve if the discriminant (b^2 - 4ac) of the quadratic equation kx^2 + (1+k)x + k = 0 is equal to zero.

Q: How do I find the values of k for which the x-axis is a tangent to the curve?

A: To find the values of k, you need to solve the quadratic equation -3k^2 + 2k + 1 = 0. You can use the quadratic formula to solve for k.

Q: What are the values of k for which the x-axis is a tangent to the curve?

A: The values of k are -1/3 and 1.

Q: What are the equations of the curve for the values of k?

A: For k = -1/3, the equation of the curve is f(x) = (-1/3)x^2 + (2/3)x - 1/3. For k = 1, the equation of the curve is f(x) = x^2 + 2x + 1.

Q: How do I determine the equations of the curve for the values of k?

A: To determine the equations of the curve, you need to substitute the values of k into the equation f(x) = kx^2 + (1+k)x + k.

Q: What is the significance of the x-axis being a tangent to the curve?

A: The x-axis being a tangent to the curve means that the curve intersects the x-axis at a single point. This is an important concept in mathematics and has many applications in various fields.

Q: How do I apply the concept of tangency to solve problems involving quadratic curves?

A: To apply the concept of tangency, you need to use the discriminant (b^2 - 4ac) of the quadratic equation and solve for the values of k. You can then use these values to determine the equations of the curve.

Q: What are some real-world applications of the concept of tangency?

A: The concept of tangency has many real-world applications, including:

Physics: The concept of tangency is used to describe the motion of objects and the forces acting on them.
Engineering: The concept of tangency is used to design and optimize systems, such as bridges and buildings.
Computer Science: The concept of tangency is used in computer graphics and game development to create realistic and interactive environments.

Q: How do I learn more about the concept of tangency and quadratic curves?

A: You can learn more about the concept of tangency and quadratic curves by:

Reading books and online resources on mathematics and physics.
Taking online courses and tutorials on mathematics and computer science.
Practicing problems and exercises to develop your skills and understanding.

Conclusion

In this article, we have answered some frequently asked questions about the concept of tangency and quadratic curves. We have discussed the condition for the x-axis to be a tangent to the curve, the values of k for which the x-axis is a tangent, and the equations of the curve for the values of k. We have also provided some real-world applications of the concept of tangency and some resources for learning more about the topic.