# Thread: parabola-hyperbola tangent point (a derivatives starting point)

1. ## parabola-hyperbola tangent point (a derivatives starting point)

Hello

I have this problem I'm trying to solve without much success, it says:
The parabola y=c-x^2 is tangent to the hyperbola y=2/x.
Draw and find the tangent point and the corresponding parameter c.

Do I find the derivative of the hyperbola and then assume it is also the derivative of the parabola? (using some equation like that to find c at the tangent point of the two?)
Not sure how to do it...
How do I begin to solve this? (I dont remember anything about parabolas and hyperbolas)

Thanks!

2. ## Re: parabola-hyperbola tangent point (a derivatives starting point)

Yes, you're on the right track. Set the derivatives (slopes) of the two equations equal, and find what the corresponding value for x is. Then set the two equations equal to each other at that value of x (since you know they go through the same point where the slopes are the same) and solve for C.

3. ## Re: parabola-hyperbola tangent point (a derivatives starting point)

I get the derivative of c-x^2 is -2x
and y'(derivative) of 2/x is -2/x^2

i get x^3 = 1 which is x=1

so if i set the two equations equal to each other i get c=3. the tangent point is (1,2). it that right?

Next i need to draw it, i wonder if it's bad to use graphing calculator for that... as i dont remember at all how to draw these..(especially hyperbolas)
and to explain what would happen if c is larger or smaller than the value we found -> according to the graphing calculator (and just intuition), if c is larger or smaller then there won't a tangent point because the
parabola won't be tangent to the hyperbola at any other point. it would basically "move" (the term is "translation" in 3D graphics?) up or down on the axis when c is larger or smaller accordingly. would that be a stratifying explanation?

Thanks again!

4. ## Re: parabola-hyperbola tangent point (a derivatives starting point)

Originally Posted by ryu1
i get c=3. the tangent point is (1,2). it that right?
Correct!

Originally Posted by ryu1
Next i need to draw it, i wonder if it's bad to use graphing calculator for that... as i dont remember at all how to draw these..(especially hyperbolas)
The graphing calculator will do a good job at showing how these curves look, but you should practice plotting the points on graph paper yourself. See what values for y you get for x = -3, -2, -1, 0, +1, +2, and +3, and connect the dots.

Originally Posted by ryu1
and to explain what would happen if c is larger or smaller than the value we found
Changing 'C' makes the parabola (which looks like an upside down cup) slide up and down the y axis. The larger the value of C the higher up the y-axis the parabola moves.

Originally Posted by ryu1
according to the graphing calculator (and just intuition), if c is larger or smaller then there won't a tangent point because the
parabola won't be tangent to the hyperbola at any other point. it would basically "move" (the term is "translation" in 3D graphics?) up or down on the axis when c is larger or smaller accordingly. would that be a stratifying explanation?
Don't know if it's "stratifying" - but I think it's satisfactory.

5. ## Re: parabola-hyperbola tangent point (a derivatives starting point)

Originally Posted by ebaines
Correct!

The graphing calculator will do a good job at showing how these curves look, but you should practice plotting the points on graph paper yourself. See what values for y you get for x = -3, -2, -1, 0, +1, +2, and +3, and connect the dots.
Oh my god, genius! I am so ashamed I didn't think of that while it's the obvious way to graph anything... great idea! (I acutally started googling for a way to graph hyerbolas and found this on youtube, i wonder if it would help Step-by-Step Drawing Tips : How to Draw a Hyperbola - YouTube

I just did it and it also refreshed the idea of limits, because we see as x grows really large then the graph approaches 0, and as it gets really small it approaches infinity (and the function isn't defined at x=0 of course)
same for the negative part.

thanks a lot!

Got 95/100 points on my last work (also) thanks to this forum.