1. Derivatives and Perpendicular Lines

Determine the coordinates of the point of intersection of the two perpendicular lines that intersect on the y-axis and are both tangent to the parabola given below:

$y = 9x^2$

When I sketch this out, I can see visually that this point of intersection cannot occur at a positive y-value. Other things that are quickly apparent:

$f'(x) = 18x$
The point of intersection must have an x-coordinate of 0.

So, if I use the formula of a line, and plug in an x-coordinate of 0, I get:

$y = f(0) + f'(0)(x - 0)$
$y = 0 + 0(x - 0)$
$y = 0$

And I conclude that my point of intersection is (0,0). All fine and dandy, except that this isn't the right answer. Can anybody help?

Thanks.

2. The two tangents should be at right angles to each other, and at the same time tangent to the curve.

Let the gradient of the first line be $m_1$

That of the second will be $m_2$

And $m_1 \times m_2 = -1$

3. Yeah, I thought of that (and probably should have listed it in my original post), but I still don't know how to integrate it into this solution. Here's some further thoughts:

Let $L_1$ = the positively sloping line and $L_2$ be the negatively sloping line.

Let $L_1$ pass through the points, $(x_1, y_1)$ and $(0, y_2)$ where $y_2 < 0$.

Since $y = 9x^2$ is even, this means that $L_2$ passes through $(-x_1, y_1)$ and $(0, y_2)$.

As you point out, the slopes are negative reciprocals of each other, by definition.

Then, there are a couple of things that I think are true, but I don't know just how to prove them. First $m_1 = 1$ and $m_2 = -1$. I say this because, in order to be tangent to an even function, the only way that I can see two perpendicular lines meeting at the y-axis would be if they both had symmetrical x-coordinates (e.g. (5, 0) and (-5, 0)).

If that is true, this final one might be the key to the whole thing. If the slope is 1 or -1, and the lines intersect at the y-axis, then doesn't $y_2$ have to equal $-(y_1)$?

Am I on the right track? Making this to complicated? Any help is appreciated.

Thanks again.

4. That's right! m1 and m2 equal to -1 and 1 (order unimportant).

So that, the gradient at anypoint being 18x, you get:

$18x = 1$

$x = \dfrac{1}{18}$

And:

$18x = -1$

$x = -\dfrac{1}{18}$

Those are your two x coordinates. In your picture, the points where the tangents meet the line have the same y coordinate.

At x = 1/18, find y and from the points and slope, get the equation of the lines (in fact, one only is necessary)

Then, find where the line meets the y axis.

5. Originally Posted by joatmon
Yeah, I thought of that (and probably should have listed it in my original post), but I still don't know how to integrate it into this solution. Here's some further thoughts:

Let $L_1$ = the positively sloping line and $L_2$ be the negatively sloping line.

Let $L_1$ pass through the points, $(x_1, y_1)$ and $(0, y_2)$ where $y_2 < 0$.

Since $y = 9x^2$ is even, this means that $L_2$ passes through $(-x_1, y_1)$ and $(0, y_2)$.

As you point out, the slopes are negative reciprocals of each other, by definition.

Then, there are a couple of things that I think are true, but I don't know just how to prove them. First $m_1 = 1$ and $m_2 = -1$. I say this because, in order to be tangent to an even function, the only way that I can see two perpendicular lines meeting at the y-axis would be if they both had symmetrical x-coordinates (e.g. (5, 0) and (-5, 0)).

If that is true, this final one might be the key to the whole thing. If the slope is 1 or -1, and the lines intersect at the y-axis, then doesn't $y_2$ have to equal $-(y_1)$?

Am I on the right track? Making this to complicated? Any help is appreciated.

Thanks again.
you are on the right track ... the use of symmetry is key to this problem.

6. Got it. Thanks to both of you for your help.

$f(\frac{1}{18}) = 9(\frac{1}{18})^2 = \frac{1}{36}$

So the tangental points are $(\frac{1}{18}, \frac{1}{36})$ and $(\frac{-1}{18}, \frac{1}{36})$

And that means that the lines intersect at $(0, \frac{-1}{36})$.

Can't thank you enough for your help.
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