1. ## Very confused...(derivatives)

I am working on some work that was assigned to me in my calculus class and I don't understand how I am supposed to start this problem. Can anyone guide me in the right direction or get me started off?

Here's a link to the problem: ImageShack&#174; - Online Photo and Video Hosting

Not asking for anyone to do it for me, I just need some help getting started .

Thanks

2. let a point on the parabola's upper branch be $(x,y)$ ... note that symmetry will take care of the normal line on the lower branch

slope between $(x,y)$ and $(a,0)$ ...

$m = \dfrac{y}{x-a}$

slope of the upper branch at any point $(x,y)$ ... $y' = \dfrac{1}{2y}$

slope normal to any point $(x,y)$ ... $-\dfrac{1}{y'} = -2y$

note that the normal slope equals the slope between $(x,y)$ and $(a,0)$ and that $x > 0$

set up an equation, solve for $x$ in terms of $a$ and then sub into the inequality $x > 0$

3. ## would this answer suffice?

I worked on this for a while last night and came up with this for an answer...have I answered it correctly?

3a. Let c > 0 be a point along the positive real axis. Then the tangent to the parabola at (c,sqrt(c)) has slope
1/(2sqrt(c)) => the normal line has slope -2sqrt(c). point slope form: the normal line has equation
y - sqrt(c) = -2sqrt(c)(x - c)

this intercepts the x-axis where y = 0 => -sqrt(c) = -2sqrt(c)(a-c) => a = 1/2 + c.

therefore a must be at least 1/2.

3b. the normal lines are perpendicular when they have pi/4 angles to the real axis.
this occurs when -2sqrt(c) = -1 => c = 1/4 => a = 3/4.

4. That looks good to me.