Consider a parabola y = x² .
The line that goes through the point (0,3/2) and is orthogonal to a tangent line to the part of parabola y=x² with x0 is

y= -x/2 + 3/2 ,
and x-coordinate of the intersection of the above two lines is [??]

Question

Consider a parabola y = x² .
The line that goes through the point (0,3/2) and is orthogonal to a tangent line to the part of parabola y=x² with x>0 is

y= -x/2 + 3/2 ,
and x-coordinate of the intersection of the above two lines is [??]

styxer · Answer

So, to found the  -1/2 I derivated the y=x² y'=2x and did m1 * m2 = -1 , applied in
 (y-3/2)= -1/2 * (x-0) to find y = -x/2 + 3/2 ...

But calculating 2x = -x/2 + 3/2 , I don't find the correct answer of x-coordinate... why? :(

stamp · Answer

|dw:1479524991288:dw|

loser66 · Answer

because the intersection point between y =x^2 and the line -x/2 + 3/2, not 2x with -x/2+3/2

styxer · Answer

@Loser66 ok but the question are saying "two lines", shouldn't we assume that these lines are what we just found?

loser66 · Answer

common!! " line"  is not used only for a straight one.

styxer · Answer

@Loser66 oh... doing x² = -x/2 + 3/2 I found the answer...

loser66 · Answer

Tell me, what do you call the latitude line? a curve or a line?

loser66 · Answer

ok, good

styxer · Answer

@Loser66 sorry, sometimes I get confused about these terms... English is not my native language ^_^'

styxer · Answer

thanks for your help

loser66 · Answer

neither mine.:)