anonymous
  • anonymous
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...
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
is this a calculus question?
anonymous
  • anonymous
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.
anonymous
  • anonymous
how do you figure

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anonymous
  • anonymous
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.
anonymous
  • anonymous
so the slope of the normal line is -1/(2x)??
anonymous
  • anonymous
yep
anonymous
  • anonymous
i thought the slope had to be a number
anonymous
  • anonymous
like so that's equal to a slope of -1/2?
anonymous
  • anonymous
yes but the slope on a parabola depends on what point you look at. so the slope is given as an expression.
anonymous
  • anonymous
ohh for real ok
anonymous
  • anonymous
did you get to this point of the problem: y = c- 1/2?
anonymous
  • anonymous
i don't know i mean y-c = -(1/2x)(x-0)? or -(1/2)(x-0) ???
anonymous
  • anonymous
...the second right
anonymous
  • anonymous
yes... you got it now just simplify the right side to -1/2*x
anonymous
  • anonymous
right obviously so ok then yeah y = c-1/2*x
anonymous
  • anonymous
sorry, i mean -1/2
anonymous
  • anonymous
no x's
anonymous
  • anonymous
wait what
anonymous
  • anonymous
simplifying the right side, the x's cancell out right? you're left with only -1/2
anonymous
  • anonymous
where'd the x go
anonymous
  • anonymous
oh ok so it IS -1/2*x(x-0) right to begin with
anonymous
  • anonymous
just to be more clear, the right side is (-1/(2x))*(x-0)
anonymous
  • anonymous
right right that's what i meant ok
anonymous
  • anonymous
thanks
anonymous
  • anonymous
np
anonymous
  • anonymous
so if c = 1/2, y = 0, and... how does that mean that there is one line and that it's the y-axis?
anonymous
  • anonymous
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
anonymous
  • anonymous
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?
anonymous
  • anonymous
from your equation, y=x^2, notice that x^2 is always positive... so y must always be positive.
anonymous
  • anonymous
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....
anonymous
  • anonymous
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
anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
oh they're each other points on the normal line that goes through the point (0,c) so that makes a line
anonymous
  • anonymous
one for each result i got it sorry
anonymous
  • anonymous
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...
anonymous
  • anonymous
gotcha
anonymous
  • anonymous
nice problem... thank you
anonymous
  • anonymous
aw no thank you

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