## calyne 3 years ago If c > (1/2), how many lines through the point (0,c) are normal lines to the parabola y = x^2? What if c < or = (1/2)? Please explain the steps...

1. dpaInc

is this a calculus question?

2. dpaInc

if c=1/2 then there is only one line... that is the y-axis. if c<1/2, there are no lines because the parabola will always be positive. Now if c>1/2, there are always two lines.

3. calyne

how do you figure

4. dpaInc

take the derivative of y=x^2 to get y'=2x. This gives the slope of the tangent line for what ever x coordinate you look at that's on the parabola. Since you want the line that's normal to the parabola, take the negative reciprocal of the derivative to get you the slope of the line perpendicular to the tangent line. Since you know the line goes through (0, c) you can use that as the point to get the equation of the line perpendicular to tangent line using point-slope form of a line. you should end up with this equation: y = c - 1/2. this is how I came to the conclusions in my previous post.

5. calyne

so the slope of the normal line is -1/(2x)??

6. dpaInc

yep

7. calyne

i thought the slope had to be a number

8. calyne

like so that's equal to a slope of -1/2?

9. dpaInc

yes but the slope on a parabola depends on what point you look at. so the slope is given as an expression.

10. calyne

ohh for real ok

11. dpaInc

did you get to this point of the problem: y = c- 1/2?

12. calyne

i don't know i mean y-c = -(1/2x)(x-0)? or -(1/2)(x-0) ???

13. calyne

...the second right

14. dpaInc

yes... you got it now just simplify the right side to -1/2*x

15. calyne

right obviously so ok then yeah y = c-1/2*x

16. dpaInc

sorry, i mean -1/2

17. dpaInc

no x's

18. calyne

wait what

19. dpaInc

simplifying the right side, the x's cancell out right? you're left with only -1/2

20. calyne

where'd the x go

21. calyne

oh ok so it IS -1/2*x(x-0) right to begin with

22. dpaInc

just to be more clear, the right side is (-1/(2x))*(x-0)

23. calyne

right right that's what i meant ok

24. calyne

thanks

25. dpaInc

np

26. calyne

so if c = 1/2, y = 0, and... how does that mean that there is one line and that it's the y-axis?

27. dpaInc

when y=0, this makes x=0. so the line normal to the parabola at (0,0) is the vertical line, x=0, and this is the only one when c=1/2

28. calyne

oh.. k. so if c < 1/2, it's negative, and x^2 is always positive on y... how would i check that otherwise though if i didn't know the graph of the function?

29. dpaInc

from your equation, y=x^2, notice that x^2 is always positive... so y must always be positive.

30. calyne

but uh i'm sorry i know i must seem stupid as hell but how does that (c = 1/2, y = 0, x = 0, etc) mean that x is always 0 on the line? that's not just a point? geez....

31. calyne

that i know i was asking like if i didn't know the graph of the function like i know the graph of x^2... but obviously it's no big deal, same deal, it's gotta fit in the domain/range..... but uh there i'm a little fuzzy anyway what the hell is c in the normal line

32. dpaInc

it does not mean that x is always zero, all I did was to solve y=x^2 when y=0. Now, if c>1/2, notice that your y is positive so that yields two x values, one positive and the other negative.

33. calyne

alright but so what is that you're finding? how is the resulting x,y a LINE and not a POINT? you know what i mean? that's how i'm confused

34. calyne

oh they're each other points on the normal line that goes through the point (0,c) so that makes a line

35. calyne

one for each result i got it sorry

36. dpaInc

remember the question was "how many lines..." ? the resulting points gave you the other point(s) needed to connect with (0, c) and be normal to the parabola...

37. calyne

gotcha

38. dpaInc

nice problem... thank you

39. calyne

aw no thank you