Problem 52
Question
Find \(k\) if the curve \(y=x^{2}+k\) is tangent to the line \(y=2 x\)
Step-by-Step Solution
Verified Answer
The value of \(k\) that makes the curve tangent to the line is \(k = 1\).
1Step 1: Understand the Problem
We are asked to find the constant \(k\) for the quadratic curve \(y = x^2 + k\) so that it is tangent to the line \(y = 2x\). Being tangent means the curve and the line touch at exactly one point, and at that point, their slopes are equal.
2Step 2: Set the Equations Equal
To find the point of tangency, set \(x^2 + k = 2x\). Rearranging this gives the equation \(x^2 - 2x + k = 0\).
3Step 3: Use the Tangency Condition
Since the line is tangent to the curve, the quadratic equation \(x^2 - 2x + k = 0\) must have exactly one solution. For this to happen, the discriminant \(b^2 - 4ac\) of the quadratic equation must be zero. Here, \(a = 1\), \(b = -2\), and \(c = k\).
4Step 4: Solve for k Using the Discriminant
Set the discriminant \((-2)^2 - 4\times1\times k = 0\). Solving this, we get \(4 - 4k = 0\). Therefore, \(4k = 4\), and solving for \(k\), we find \(k = 1\).
5Step 5: Verify the Solution by Checking Slope
Check that the slope of the curve at the point of tangency matches the slope of the line \(y = 2x\). The derivative of \(y = x^2 + 1\) is \(y' = 2x\). Set \(2x = 2\), which is consistent for \(x = 1\). Substituting \(x = 1\) into the curve equation gives \(y = 1^2 + 1 = 2\), matching the line \(y = 2\).
Key Concepts
Quadratic CurveDiscriminantDerivative
Quadratic Curve
A quadratic curve, often represented as a parabola, is a type of polynomial curve defined by a quadratic equation. This equation is typically of the form \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. In our exercise, the quadratic curve is given as \(y = x^2 + k\).
This specific curve is simplified because it lacks a linear \(b\) term, which means the parabola is symmetrical around the y-axis if translated appropriately. The vertex of this parabola is positioned at the point \( (0, k) \), as derived from its standard form.
When a line is tangent to this curve, it touches it at exactly one point without crossing over. To achieve this, both the curve's position and its direction at the point of tangency must be precisely adjusted. This is where the concepts of the discriminant and derivative become useful.
This specific curve is simplified because it lacks a linear \(b\) term, which means the parabola is symmetrical around the y-axis if translated appropriately. The vertex of this parabola is positioned at the point \( (0, k) \), as derived from its standard form.
When a line is tangent to this curve, it touches it at exactly one point without crossing over. To achieve this, both the curve's position and its direction at the point of tangency must be precisely adjusted. This is where the concepts of the discriminant and derivative become useful.
Discriminant
The discriminant is a key algebraic tool used to determine the nature of the roots of a quadratic equation. For a standard quadratic equation \(ax^2 + bx + c = 0\), the discriminant is calculated as \(b^2 - 4ac\).
In the context of our exercise, determining the tangency between the line and the quadratic curve hinges on the discriminant being zero. This condition means that the quadratic curve \(x^2 - 2x + k = 0\) has exactly one solution, signifying a single point of contact.
Let's break down the steps:
In the context of our exercise, determining the tangency between the line and the quadratic curve hinges on the discriminant being zero. This condition means that the quadratic curve \(x^2 - 2x + k = 0\) has exactly one solution, signifying a single point of contact.
Let's break down the steps:
- Here, we identify \(a=1\), \(b=-2\), and \(c=k\) in our quadratic equation.
- The discriminant formula gives \((-2)^2 - 4 imes 1 imes k\).
- For the line to be tangent, set the discriminant to zero: \(4 - 4k = 0\).
- Solving \(4k = 4\) ensures \(k = 1\).
Derivative
To fully confirm the tangency condition, examining the derivative of the quadratic curve is crucial. The derivative indicates the slope of the curve at any given point.
For the function \(y = x^2 + k\), the derivative, \(y'\), is calculated as follows:\[y' = \frac{d}{dx}(x^2 + k) = 2x.\]
This derivative implies that at any point \(x\), the slope of the curve is \(2x\).
To verify the tangency condition:
Thus, the derivative helps confirm that at \(x=1\), the slope of the curve is equal to the slope of the line, supporting the tangency assertion.
For the function \(y = x^2 + k\), the derivative, \(y'\), is calculated as follows:\[y' = \frac{d}{dx}(x^2 + k) = 2x.\]
This derivative implies that at any point \(x\), the slope of the curve is \(2x\).
To verify the tangency condition:
- At the point of tangency, our line has a constant slope of 2, represented by its equation \(y = 2x\).
- Equating the slope of the curve to the slope of the line requires setting \(2x = 2\).
- Solving gives \(x = 1\).
Thus, the derivative helps confirm that at \(x=1\), the slope of the curve is equal to the slope of the line, supporting the tangency assertion.
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