Proving a tangent line of a parabola

joejoe1

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Here is the problem my Geometry textbook asks me to prove:
a tangent line of a parabola is a line that intersects but does not cross the parabola. Prove that a line tangent to the parabola x^2=4py at the point (a,b) crosses the y-axis at (0,-b).

From that I can draw the parabola up and down and the line on a given point (a,b) but really don't know how to start. Any help anyone could provide to help me get started would be appreciated.
 
Here is the problem my Geometry textbook asks me to prove:
a tangent line of a parabola is a line that intersects but does not cross the parabola. Prove that a line tangent to the parabola x^2=4py at the point (a,b) crosses the y-axis at (0,-b).

From that I can draw the parabola up and down and the line on a given point (a,b) but really don't know how to start. Any help anyone could provide to help me get started would be appreciated.
You have two points on the line, namely (a,b) and (0,-b), so get the equation of this line. Now verify what you are being asked. If you need more help that tell us what you did so far. BTW, this is a great problem.
 
So I think I found the equation of the line made by the two points but I guess I'm still not sure how to prove that its tangent to the parabola. (we haven't really covered tangent lines too much yet). Here is the process I went through to get the equation of the line.
given (a,b) and (0,-b):
slope=2b/a
point slope form: y--b=(2b/a)(x) --> y+b=(2b/a)(x)
 
So I think I found the equation of the line made by the two points but I guess I'm still not sure how to prove that its tangent to the parabola. (we haven't really covered tangent lines too much yet). Here is the process I went through to get the equation of the line.
given (a,b) and (0,-b):
slope=2b/a
point slope form: y--b=(2b/a)(x) --> y+b=(2b/a)(x)
OK, so you found the equation of the line is y=(2b/a)(x)-b. Now you need to show that the parabola is always above the line if p>0. Graph both the parabola and the line and SEE that this is true. You should then consider a new function, y=[x^2/(4p)] - [(2b/a)(x)-b] and show that it is always positive. Since it is a quadratic this should be simple. Then do this all over for p<0.
Show us your work and we will assist further.
 
I'm having issues proving that [x^2/(4p)] - [(2b/a)(x)-b] is > 0. You mentioned that it is a quadratic and that that should somehow make it simple?! :'(. Heres what I've tried doing the past half an hour:

Because I'm trying to prove that
[x^2/(4p)] - [(2b/a)(x)-b] is > 0, I set the equation such that [x^2/(4p)] - [(2b/a)(x)-b] > 0. Then because I remembered to solve an inequality you exchange the inequality with an = sign to solve, I did that next. (Maybe that's where I went wrong?) When I got to this point I didn't really know what I was solving for on either side, or even what I was doing. Maybe I went about proving that entirely the wrong way? I don't really know. I also briefly entertained the idea of using the quadratic equation or factoring but couldn't see how that would prove anything. Ahhhh! I really wish I was better at proofs!
 
I'm having issues proving that [x^2/(4p)] - [(2b/a)(x)-b] is > 0. You mentioned that it is a quadratic and that that should somehow make it simple?! :'(. Heres what I've tried doing the past half an hour:

Because I'm trying to prove that
[x^2/(4p)] - [(2b/a)(x)-b] is > 0, I set the equation such that [x^2/(4p)] - [(2b/a)(x)-b] > 0. Then because I remembered to solve an inequality you exchange the inequality with an = sign to solve, I did that next. (Maybe that's where I went wrong?) When I got to this point I didn't really know what I was solving for on either side, or even what I was doing. Maybe I went about proving that entirely the wrong way? I don't really know. I also briefly entertained the idea of using the quadratic equation or factoring but couldn't see how that would prove anything. Ahhhh! I really wish I was better at proofs!

Since the (a, b) is on the parabola it must be that
a^2=4pb
or
4pb - a^2= 0
How is that equation related to the discriminate of the quadratic? And, what does that say about the quadratic itself changing signs.
 
I'm having issues proving that [x^2/(4p)] - [(2b/a)(x)-b] is > 0. You mentioned that it is a quadratic and that that should somehow make it simple?! :'(. Heres what I've tried doing the past half an hour:

Because I'm trying to prove that
[x^2/(4p)] - [(2b/a)(x)-b] is > 0, I set the equation such that [x^2/(4p)] - [(2b/a)(x)-b] > 0. Then because I remembered to solve an inequality you exchange the inequality with an = sign to solve, I did that next. (Maybe that's where I went wrong?) When I got to this point I didn't really know what I was solving for on either side, or even what I was doing. Maybe I went about proving that entirely the wrong way? I don't really know. I also briefly entertained the idea of using the quadratic equation or factoring but couldn't see how that would prove anything. Ahhhh! I really wish I was better at proofs!
Parabolas have a max or min at x=-b/(2a). I was hoping that you knew this. If the min(max) value is positive(negative) then the function is always positive(negative)
 
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