Introduction

In the field of algebra, finding the tangent line to a curve at a specific point can be done through algebraic methods without resorting to calculus. This article demonstrates how to find the slope of the tangent to the parabola (y 4x^2 - 5x - 2) at a specific point, and how to determine the point on the parabola where the tangent is parallel to a given line. Let's explore the steps involved in this process.

Step 1: Finding the Slope of the Tangent at a Specific Point on the Parabola

Given the equation of the parabola (y 4x^2 - 5x - 2).

To find the slope of the tangent at any point on the parabola, we start by recognizing that the slope can be derived from the equation of the parabola. In this case, the slope of the tangent line at any point (x a) on the parabola is given by the derivative of the equation with respect to (x).

The derivative of (y 4x^2 - 5x - 2) is:

[m frac{dy}{dx} 8x - 5]

Therefore, the slope of the tangent at (x a) is:

[m 8a - 5]

Step 2: Determining the Point Where the Tangent is Parallel to a Given Line

Next, consider the line given by the equation:

[1 - 2y - 18 0]

Reorganizing this equation to the standard form, we get:

[1 - 18 2y implies -17 2y implies y frac{-17}{2}]

This line can be rewritten in slope-intercept form as:

[2y 1 - 18 implies 2y -17 implies y -frac{17}{2}]

However, the slope of this line is:

[frac{1}{2}]

But it seems there was a typo or misinterpretation. The correct slope for the line (10 - 2y - 18 0) can be derived from the slope form (2y 1 - 18 implies y frac{1}{2}x - 9), so the slope (m_2 frac{1}{2}). However, rewriting it correctly as (10 - 2y - 18 0 implies 2y 18 - 10 implies y 5x - 9), the slope is (m_2 5).

For the tangent to be parallel to this line, the slopes must be equal:

[8a - 5 5 implies 8a 10 implies a frac{5}{4}]

Step 3: Finding the Specific Point where the Tangent is Parallel to the Line

Substituting (a frac{5}{4}) into the equation of the parabola to find the corresponding (y)-coordinate:

[y 4left(frac{5}{4}right)^2 - 5left(frac{5}{4}right) - 2]

[ 4 cdot frac{25}{16} - frac{25}{4} - 2]

[ frac{100}{16} - frac{100}{16} - frac{32}{16} -frac{32}{16} -2]

Thus, the point on the parabola where the tangent is parallel to the line is:

((frac{5}{4}, -2))

Conclusion

In conclusion, the slope of the tangent to the parabola (y 4x^2 - 5x - 2) at (x a) is given by (8a - 5). The point on the parabola where the tangent is parallel to the line (10 - 2y - 18 0) is (left(frac{5}{4}, -2right)).