Final answer:
To find the equation of the normal line to the parabola y = x² - 8x + 7 that is parallel to the line x - 6y = 3, we need to find the negative reciprocal of the slope of the given line and the tangent line of the parabola at the point of tangency. Using the point-slope form, we can then find the equation of the normal line.
Step-by-step explanation:
To find the equation of the normal line to the parabola y = x² - 8x + 7 that is parallel to the line x - 6y = 3, we need to understand that the normal line will have a slope that is the negative reciprocal of the slope of the given line.
First, we rewrite the given line in the slope-intercept form: y = (1/6)x - (1/2).
The slope of the given line is 1/6, so the slope of the normal line will be -6 (negative reciprocal).
Next, we find the derivative of the parabola to get the slope of the tangent line. The derivative of y = x² - 8x + 7 is y' = 2x - 8.
Since the slope of the tangent line is equal to the slope of the normal line at the point of tangency, we set 2x - 8 = -6. Solving for x, we find x = 1.
Substituting x = 1 back into the original equation y = x² - 8x + 7, we get y = 0.
Therefore, the point of tangency is (1, 0).
Using the point-slope form, we can find the equation of the normal line: y - 0 = -6(x - 1). Simplifying, we get y = -6x + 6.
So, the equation of the normal line to the parabola that is parallel to the line is y = -6x + 6.