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.

Printable View

- Jan 10th 2010, 11:09 PMNaplesPerpendicular 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 PMbandedkrait
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 PMProve It
$\displaystyle F'(x) = 2kx$.

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

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

Since the tangent lines are perpendicular,

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

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

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

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

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