Find an equation of the normal line to the parabola y = x2 − 6x + 5 that is parallel to the line x − 4y = 6.

Question

Find an equation of the normal line to the parabola y = x² − 6x + 5

 that is parallel to the line x − 4y = 6.

y=_______

Answer 1

Since the normal line is to be parallel to the given line, we need to determine the slope of the given line. Switching the equation x - 4y = 6 over to slope-intercept form, we have y = 1/4 x - 3/2 so we see that the slope of the given line, and thus of our normal line, is 1/4.

Since the slope of the normal line is 1/4, this means that it must intersect the parabola to form a right angle, and thus the slope of the parabola at the point where the line hits it must be its opposite reciprocal, or -4. So now we take the derivative of the given quadratic equation, which is f '(x) = 2x - 6 and we set this equal to what its slope should be, which is -4. Solving 2x - 6 = -4, we have x = 1. Plugging in this x value into the original quadratic equation, we have y = 1² - 6(1) + 5 or y = 0. So the point on the parabola where the normal line would hit it to have the correct slope is ( 1 , 0 ).

Now, using the slope of the normal line and the point on the parabola where it must make contact with the parabola, we can write the equation in point-slope form and then switch it to slope-intercept form:

y - 0 = 1/4( x - 1 ). Switching this over to slope-intercept form, we have y = 1/4 x - 1/4.