Question

Find an equation of the normal line to the parabola y = x^2 − 8x + 9 that is parallel to the line x − 2y = 3?

Answer 1

`y = 1/2x -15/2`

Explanation:

We have a parabola with the equation:

`y = x^2-8x+9`

Differentiating wrt `x` we have:

`dy/dx = 2x - 8`

If we consider the line:

`x-2y =3`

and putting into the standard form, `y=mx+c`, we have:

`2y=x-3 => y =1/2x-3/2`

So, the given line has gradient, `m = 1/2`

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. The normal is perpendicular to the tangent. So, the normal should have the same gradient, `m`

As the product of the gradients of perpendicular lines is `-1`, then we require that:

`2x-8 = -1/m => 2x = 8-2 => x=3`

So the point that satisfies the question, has `x`-coordinate, `x=3`

`x=3 => y =9-24+9 = -6`

The equation we seek passes through `(3,-6)` and has slope `1/2`. Using the point/slope form `y-y_1=m(x-x_1)` the normal equations we seek is;

`y- (-6)=(1/2)(x-3)`

`:. y+6 = 1/2x -3/2`

`:. y = 1/2x -15/2`

We can confirm this graphically: