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anonymous
 4 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...
anonymous
 4 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...

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is this a calculus question?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if c=1/2 then there is only one line... that is the yaxis. 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
 4 years ago
Best ResponseYou've already chosen the best response.0take 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 pointslope 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
 4 years ago
Best ResponseYou've already chosen the best response.0so the slope of the normal line is 1/(2x)??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i thought the slope had to be a number

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0like so that's equal to a slope of 1/2?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes but the slope on a parabola depends on what point you look at. so the slope is given as an expression.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0did you get to this point of the problem: y = c 1/2?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i don't know i mean yc = (1/2x)(x0)? or (1/2)(x0) ???

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes... you got it now just simplify the right side to 1/2*x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0right obviously so ok then yeah y = c1/2*x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0simplifying the right side, the x's cancell out right? you're left with only 1/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh ok so it IS 1/2*x(x0) right to begin with

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just to be more clear, the right side is (1/(2x))*(x0)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0right right that's what i meant ok

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so if c = 1/2, y = 0, and... how does that mean that there is one line and that it's the yaxis?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0when 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
 4 years ago
Best ResponseYou've already chosen the best response.0oh.. 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
 4 years ago
Best ResponseYou've already chosen the best response.0from your equation, y=x^2, notice that x^2 is always positive... so y must always be positive.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but 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
 4 years ago
Best ResponseYou've already chosen the best response.0that 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
 4 years ago
Best ResponseYou've already chosen the best response.0it 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
 4 years ago
Best ResponseYou've already chosen the best response.0alright 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
 4 years ago
Best ResponseYou've already chosen the best response.0oh they're each other points on the normal line that goes through the point (0,c) so that makes a line

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0one for each result i got it sorry

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0remember 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
 4 years ago
Best ResponseYou've already chosen the best response.0nice problem... thank you
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