# Question #60d5b

Sep 23, 2015

Find the tangent line to the parabola at the point x = 2, then find the normal line (perpendicular) to that tangent line at the point x = 2.

#### Explanation:

STEP 1:
Find the tangent line(s) to the parabola at x = 2. If you graph this parabola, you will see that there are actually TWO tangent lines at the point x = 2. So, there are two normal lines ... one for each tangent line.

At x = 2, ${y}^{2}$ = 4x = 4(2) = 8 [now solve for y]
y = $\pm$ $\sqrt{8}$ = $\pm$2$\sqrt{2}$

So, the points of tangency are [2, 2$\sqrt{2}$] and [2, - $2 \sqrt{2}$]

Now, each tangent line (y = mx + b) and the parabola ${y}^{2} = 4 x$ intersect at one point (the point of tangency). We can set these two equations equal to each other, then solve for the slope m. We can do this using substitution for the variable x:

y = mx + b = m[${y}^{2}$/4] + b

Now, simplify ...

m${y}^{2}$/4 - y + b = 0

This is a quadratic equation, so use the quadratic formula to solve for y:

y =$\frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(\frac{m}{4}\right) \left(b\right)}}{\left(2\right) \left(\frac{m}{4}\right)}$

Now, the discriminant (the part in the square root) must equal zero because there is only one point where the tangent line and the parabola intersect. So, this whole mess simplifies to:

y = 2/m

Finally, we can solve for m, the slope of the tangent lines because we know the values of y at the point of tangencies: (2$\sqrt{2}$ and - $2 \sqrt{2}$). Inserting these values for y into the above equation:

2$\sqrt{2}$ = 2/m {so, m = $\frac{\sqrt{2}}{2}$}
-2$\sqrt{2}$ = 2/m {so, m = -$\frac{\sqrt{2}}{2}$}

Now that we know the slopes of the "tangent" lines, we can calculate the slope of the "normal" lines.

STEP 2:
The slope of the normal line ( m' ) is always the negative reciprocal of the slope of the tangent line (m).

If m = $\frac{\sqrt{2}}{2}$, then m' = $- \frac{2}{\sqrt{2}}$ = $- \sqrt{2}$
If m = $- \frac{\sqrt{2}}{2}$, then m' = $+ \frac{2}{\sqrt{2}}$ = $+ \sqrt{2}$

STEP 3:
Calculate the two normal lines at the points of tangency:
[2, 2$\sqrt{2}$] and [2, - $2 \sqrt{2}$]

y = m'x + b

$2 \sqrt{2}$ = $- \sqrt{2} \cdot 2 + b$, so $b = 4 \sqrt{2}$
y = $- x \sqrt{2} + 4 \sqrt{2}$

The second normal line is:
$- 2 \sqrt{2}$ = $+ \sqrt{2} \cdot 2 + b$, so $b = - 4 \sqrt{2}$
y = $- x \sqrt{2} - 4 \sqrt{2}$

Hope that helped!