Show that for the curve y=x^2+2k1+k2, the chord joining the points x=a and x=b is parallel to the tangent at x=(a+b)/2

To solve this problem, we need to show that the chord joining the points \( x = a \) and \( x = b \) on the curve \( y = x^2 + 2k_1 + k_2 \) is parallel to the tangent to the curve at the point \( x = \frac{a+b}{2} \).

### Step-by-Step Solution

#### Step 1: Calculate the Slope of the Chord

The equation of the curve is given as:
\[ 
y = x^2 + 2k_1 + k_2 
\]

The points on the curve at \( x = a \) and \( x = b \) are:
- \( P(a, a^2 + 2k_1 + k_2) \)
- \( Q(b, b^2 + 2k_1 + k_2) \)

The slope of the chord \( PQ \) is calculated as follows:
\[
m_{\text{chord}} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{b^2 + 2k_1 + k_2 - (a^2 + 2k_1 + k_2)}{b - a}
\]

\[
= \frac{b^2 - a^2}{b - a}
\]

Using the algebraic identity \( b^2 - a^2 = (b-a)(b+a) \), we can simplify the slope of the chord as:
\[
m_{\text{chord}} = \frac{(b-a)(b+a)}{b-a} = b+a
\]

Therefore, the slope of the chord joining the points \( (a, y(a)) \) and \( (b, y(b)) \) is:
\[
m_{\text{chord}} = b + a
\]

#### Step 2: Calculate the Slope of the Tangent

To find the slope of the tangent at \( x = \frac{a+b}{2} \), we first need to find the derivative of \( y \) with respect to \( x \).

The function is:
\[
y = x^2 + 2k_1 + k_2
\]

The derivative \( y' \) (or \( \frac{dy}{dx} \)) is:
\[
\frac{dy}{dx} = 2x
\]

The slope of the tangent at \( x = \frac{a+b}{2} \) is:
\[
m_{\text{tangent}} = 2 \cdot \left(\frac{a+b}{2}\right) = a + b
\]

#### Step 3: Show Parallelism

For the chord to be parallel to the tangent, their slopes must be equal:
\[
m_{\text{chord}} = m_{\text{tangent}}
\]

From the above calculations:
\[
b+a = a+b
\]

Thus, the slopes are indeed equal, showing that the chord joining the points \( x = a \) and \( x = b \) is parallel to the tangent at \( x = \frac{a+b}{2} \).

### Conclusion

The chord joining the points on the curve \( y = x^2 + 2k_1 + k_2 \) at \( x = a \) and \( x = b \) is parallel to the tangent at \( x = \frac{a+b}{2} \) because both have the same slope \( a + b \).

---

### Additional Questions and Exploration:

1. **How do you find the equation of a tangent line to a curve at a specific point?**
2. **What is the geometric interpretation of the derivative of a function?**
3. **Can you derive a similar result for a cubic function?**
4. **What is the significance of parallel lines in analytic geometry?**
5. **How can this result be applied to determine symmetry in curves?**
6. **How does this relate to the concept of a midpoint in coordinate geometry?**
7. **What are some real-world applications of tangent and chord parallelism?**
8. **Can you generalize this method for any polynomial function?**

### Tip:

In problems involving slopes of tangents and chords, leveraging algebraic identities such as the difference of squares can significantly simplify the calculations. Understanding these geometric interpretations enhances problem-solving efficiency.

Explore how to demonstrate that for the curve y=x^2+2k1+k2, the chord joining points x=a and x=b is parallel to the tangent at x=(a+b)/2. Understand the relationship between slopes of chords and tangents in analytic geometry.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation