Find an equation of the normal line to the parabola

y = x2 − 7x + 5
that is parallel to the line
x − 3y = 3.

dy/dx = 2x-7 which is the slope of the tangent.

so the normal must have a slope of -1/(2x-7) or 1/(7-2x)

but that is supposed to be parallel to x-3y=3
that is, a slope of 1/3

1/(7-2x) = 1/3
3 = 7-2x
2x = 4
x = 2
when x=2, y = 4-14+5 = -5
so the normal has equation
x - 3y = c
with (2,-5) on it, so
2 + 15 = c

equation of normal is
x - 3y = 15

check: at (2,-5) slope should be -3
dy/dx = 2(2)-7 = -3
My answer for the equation of the normal is correct!!

To find the equation of the normal line to the parabola that is parallel to the line, we need to find the slope of the normal line.

The given line has the equation x − 3y = 3, which can be rewritten as -3y = -x + 3, or y = 1/3x - 1.

The slope of the given line is 1/3.

In order to find the slope of the normal line, we need to find the derivative of the parabola.

The given parabola has the equation y = x^2 - 7x + 5.

To find the derivative, we differentiate the equation with respect to x:

dy/dx = 2x - 7.

The slope of the normal line is the negative reciprocal of the derivative at the point of tangency. In other words, the slope of the normal line is the negative reciprocal of the slope of the tangent line.

Therefore, the slope of the normal line is -1/(2x - 7).

Now we can set the slope of the normal line equal to the slope of the given line and solve for x:

-1/(2x - 7) = 1/3.

To solve for x, we can cross multiply:

3 = -(2x - 7).

Simplifying, we get:

3 = -2x + 7.

Subtracting 7 from both sides, we have:

-4 = -2x.

Dividing by -2, we get:

x = 2.

Now that we have the value of x, we can substitute it into the original equation of the parabola to find the corresponding y-value:

y = (2)^2 - 7(2) + 5.

Simplifying, we get:

y = 4 - 14 + 5 = -5.

Therefore, the point of tangency on the parabola is (2, -5).

Now we have the point of tangency (2, -5) and the slope of the normal line (-1/(2x - 7)), so we can use the point-slope form of a line to find the equation of the normal line:

y - y1 = m(x - x1).

Plugging in the values, we get:

y - (-5) = -1/(2(2) - 7)(x - 2).

Simplifying, we have:

y + 5 = -1/(-3)(x - 2).

y + 5 = 1/3(x - 2).

Multiplying through by 3 to eliminate the fraction, we get:

3y + 15 = x - 2.

Rearranging the equation, we have:

x - 3y = 17.

Therefore, the equation of the normal line to the parabola that is parallel to the line x - 3y = 3 is x - 3y = 17.

To find the equation of the normal line to the parabola that is parallel to the given line, we need to follow these steps:

Step 1: Find the derivative of the given parabola.
Step 2: Calculate the slope of the normal line by finding the negative reciprocal of the derivative.
Step 3: Use the point of tangency on the parabola to find the y-intercept of the normal line.
Step 4: Write the equation of the normal line using the slope and y-intercept.

Let's go through each step in more detail:

Step 1: Find the derivative of the given parabola.
The given parabola is y = x^2 - 7x + 5. To find its derivative, we differentiate with respect to x. The derivative of x^2 is 2x, and the derivative of 7x is 7. Since the constant term 5 does not have any x term, it differentiates to 0. Therefore, the derivative of the parabola is dy/dx = 2x - 7.

Step 2: Calculate the slope of the normal line.
The slope of the normal line is the negative reciprocal of the derivative of the parabola. So, we need to find the negative reciprocal of 2x - 7. The negative reciprocal of a number is when you take its inverse and then multiply it by -1. So, the negative reciprocal of 2x - 7 is -1/(2x - 7).

Step 3: Use the point of tangency on the parabola to find the y-intercept of the normal line.
To find the y-intercept of the normal line, we need a point that lies on both the parabola and the normal line. This point is called the "point of tangency." Since we want the normal line to be parallel to the line x - 3y = 3, we can find the point of tangency by solving the system of equations formed by the parabola and the line. We substitute y = x^2 - 7x + 5 into the equation x - 3y = 3:

x - 3(x^2 - 7x + 5) = 3
x - 3x^2 + 21x - 15 = 3
-3x^2 + 20x - 18 = 0

Solving this quadratic equation will give us the x-coordinates of the point(s) of tangency.

Step 4: Write the equation of the normal line using the slope and y-intercept.
Once we have the slope (negative reciprocal) and the y-intercept (point of tangency), we can write the equation of the normal line using the point-slope form: y - y1 = m(x - x1), where (x1, y1) is the point of tangency.

By following these steps, you will be able to find the equation of the normal line to the parabola that is parallel to the given line.