# Perpendicular Tangent Lines

• Jan 10th 2010, 11:09 PM
Naples
Perpendicular Tangent Lines
F(x) = kx^2 + 3

If the tangent lines to the graph of F at (t, F(t)) and (-t, F(-t)) are perpendicular find t in terms of k.

I don't know where to start really so if someone could walk me through this I'd greatly appreciate it.
• Jan 10th 2010, 11:13 PM
bandedkrait
Definition of derivative of F(x) at a given point : It's the slope of the tangent to the graph of F(x) at that point.

So find the derivatives at t and -t, to obtain the slope of the respective tangents at the two points.

Now two lines are perpendicular if their slopes m1, m2 (say.) are such that (m1)(m2) = -1

Use this to find t in terms of k.

Note that you'll have to solve a quadratic equation in t.
• Jan 10th 2010, 11:13 PM
Prove It
Quote:

Originally Posted by Naples
F(x) = kx^2 + 3

If the tangent lines to the graph of F at (t, F(t)) and (-t, F(-t)) are perpendicular find t in terms of k.

I don't know where to start really so if someone could walk me through this I'd greatly appreciate it.

$\displaystyle F'(x) = 2kx$.

$\displaystyle F'(t) = 2kt$

$\displaystyle F'(-t) = -2kt$.

Since the tangent lines are perpendicular,

$\displaystyle F'(t)\cdot F'(-t) = -1$

$\displaystyle 2kt \cdot (-2kt) = -1$

$\displaystyle -4k^2t^2 = -1$

$\displaystyle t^2 = \frac{1}{4k^2}$

$\displaystyle t = \pm \frac{1}{2k}$.
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