# Math Help - Finding points on parabola given point on normal line

1. ## Finding points on parabola given point on normal line

Find the point(s) on the parabola f (x) = x^2 - 4x + 3 at which the normal line passes through the point (2,0).

I found the slope of the normal line: - 1/(2x-4)
And came up that one of the normal line equation is (-1/2)x + 1
But then I'm stuck. How do I find the points on the parabola. And should there be two normal lines that passes through the point or only one?
I'm only able to find one.

Any help is greatly appreciated! Thanks !!

2. Originally Posted by letzdiscuss
Find the point(s) on the parabola f (x) = x^2 - 4x + 3 at which the normal line passes through the point (2,0).

I found the slope of the normal line: - 1/(2x-4)
And came up that one of the normal line equation is (-1/2)x + 1
But then I'm stuck. How do I find the points on the parabola. And should there be two normal lines that passes through the point or only one?
I'm only able to find one.

Any help is greatly appreciated! Thanks !!
Let the point on the parabola be $(a, a^2 - 4a + 3)$. The gradient of the tangent to the parabola at this point is $m_{tangent} = 2a - 4$.

Therefore the gradient of the normal is $m_{tangent} = - \frac{1}{2a - 4} = \frac{1}{4 - 2a}$.

Therefore the equation of the normal is $y - (a^2 - 4a + 3) = \frac{1}{4 - 2a} (x - a)$.

But the point (2, 0) lies on the normal therefore

$- (a^2 - 4a + 3) = \frac{1}{4 - 2a} (2 - a)$

$\Rightarrow - a^2 + 4a - 3 = \frac{2 - a}{4 - 2a}$

Solve for $a$.

For checking:

Spoiler:
$a = 2 \pm \frac{1}{\sqrt{2}}$.