distance from origin to vertex; area; tangent line

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swimminginwork

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I dont even know where to start on this problem:
a) express the distance from the origin to the vertex of the parabola y=x^2-2bx as a function of b. note: b>0.
b) let V denote the vertex of the parabola in (a) and let A and B denote points where the curve meets the X-axis. Express the area of triangle VAB as a function of b.
c) A circle is drawn with center V, passing through A and B. A tangent line is then drawn through the smaller y-intercept of the circle. Express the x-intercept of this tangent line as a function of b.

I truly am dumbfounded on this one. any help would be appreciated!
 
For the first part.

\(\displaystyle \L\\\frac{d}{dx}[x^{2}-2bx]=2x-2b\)

\(\displaystyle 2x-2b=0, \;\ x=b\)

Sub this back into your parabola equation:

\(\displaystyle \L\\y=b^{2}-2b^{2}=-b^{2}\)

Distance formula:

\(\displaystyle \L\\\sqrt{(b-0)^{2}+(-b^{2})^{2}}=|b|\sqrt{b^{2}+1}\)

For instance, the distance from the vertex if b=1, \(\displaystyle x^{2}-2x\)

would be \(\displaystyle \L\\\sqrt{2}\)
 
thanks so much for helping! the first part, that's using derivatives, right? Do you know of any way to do this problem without using derivatives? It's summer preparation for Calculus, so we don't know anything about derivatives haha.
but thanks again!
 
swimminginwork said:
Do you know of any way to do this problem without using derivatives? It's summer preparation for Calculus, so we don't know anything about derivatives !
Click to expand...
Yes, is can be done without calculus. Lookup in precaculus 'How to find the vertex of a parabola'.
 
pka said:
swimminginwork said:
Do you know of any way to do this problem without using derivatives? It's summer preparation for Calculus, so we don't know anything about derivatives !
Click to expand...
Yes, is can be done without calculus. Lookup in precaculus 'How to find the vertex of a parabola'.
Click to expand...
Yeah, that's how I was initially trying to solve it- using the -b/2a formula, right? But for some reason this problem is just really tripping me up. I'm not sure how to express it in terms of b.
It's one of those problems that I sort of felt like I knew how to do until I tried to do it..
 
What is the co-ordinate of the vertex?

A = 1

B = -2b

h = (b)^2 - 2b*b = -b^2

so the co-ordinate of the vertex is (b, -b^2)

Co-ordinate of the origin is (0,0)

Find the distance.
 
Given the parabola \(\displaystyle y = px^2 + qx + r\quad \left( {p \not= 0} \right)\) the vertex is \(\displaystyle \left( {\frac{{ - q}}{{2p}},\frac{{ - q^2 }}{{4p}} + r} \right).\)
Now in this case: \(\displaystyle p = 1,\quad q = - 2b,\quad \& \quad r = 0.\)
 
thanks a lot for all of the help, everyone. I figured out the first part. I just saw that the last few responses were exactly what had just clicked for me. I really appreciate the help! Now I'm working on the next parts- hopefully getting started was just what I needed haha.
 
Here is a start: the two x-intercepts are (0,0) & (2b,0).
 
I don't know if anyone can confirm this, but should the answer for part b (the final area) be
1/2 b^2 [root(2b^2)]
 
swimminginwork said:
I don't know if anyone can confirm this, but should the answer for part b (the final area) be 1/2 b^2 [root(2b^2)]
Click to expand...
By no means is that correct.
The vertex is \(\displaystyle \left( {b, - b^2 } \right)\) thus the altitude is \(\displaystyle b^2\) and the base is \(\displaystyle 2 b\).
 
I got that the 2 x-intercepts are (0,0) and (2b,0)
I see your previous post had (0,2b) as the second one, so maybe that's where the discrepancy is? Do I have it mixed up? Wouldn't the "2b" have to be the x-coordinate if it's the x-intercept in question?
Anyway, I used those two coordinates and the distance formula to find the length of the base and came up with root(2b^2)
 
Alright, I just went back and saw that I made a careless math error when calculating the area. story of my life *sigh*
Thanks for all of your help with part b!
 
For part c, you have all the coordinates you need.

Here is a circle graph using b=1 as an example. I am not sure what you mean by the smaller of the y-intercepts. One will always be the origin. It should be easy to see what the other y-intercept will be in terms of b.

circlewo1.jpg
 
I'm still not quite sure how to do part c...
I've sketched it out, but I'm not sure how I would be able to find the x-intercept of the tangent line...
 
Start with assigning co-ordinate to V and radius to the circle.

Let V be (h,k) and radius = r

Then describe the circle:

(x-h)^2 + (y-k)^2 = r^2

Now find the y intercepts (x = 0)

and continue....
 
hmm. I worked it out [well, at least I think I did...] & came up with the x-intercept being equal to 2b. But it should be a negative, right? So I must have made a mistake somewhere along the way, correct?
Geez, I don't think I've ever had so much trouble with one problem. this is ridiculous.
 
Please show your work - if you want us to catch your mistake.
 
Did you look at the graph in my last post?. You're posts suggest you haven't. Your original post asked for the y-intercept. Now you want the x-intercept?.
 
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