yb1996
  • yb1996
The normal line through a point P (not through the origin) on the parabola y = x^2 will intersect the parabola at another point Q. Find the coordinates of the point P for which the distance from P to Q is minimized.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
and what are your thoughts?
yb1996
  • yb1996
I now that I need to have the points P and Q on x^2. I also know that I need an equation and then need to differentiate and set the differentiated equation to zero so that I can get the minimized value. I don't know how to obtain the initial equation though.
amistre64
  • amistre64
we need a line .... and a line requires a slope and a point. we have a point, how can we determine the slope at that point?

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yb1996
  • yb1996
we take the derivative at that point
amistre64
  • amistre64
correct, and the slope that is perpendicular to a tangent, is called normal .... given a tangent of y'(a), the normal slope is -1/y'(a) does that make sense? what is our point when x=a? what is our normal slope when x=a?
yb1996
  • yb1996
yes that makes sense. So the point at x = a would be (a,a^2) and the normal slope would be -1/(2a), right?
amistre64
  • amistre64
now we have a line y= -1/(2a) (x-a) +a^2 we have the parabola:y=x^2 they meet at x=a, and x= the other point. how do we find that other point?
yb1996
  • yb1996
we set that equation equal to x^2 ?
amistre64
  • amistre64
yep x^2 = -1/(2a)(x-a)+a^2 x^2 + 1/(2a)(x-a) = a^2 x^2 + 1/(2a)(x-a) - a^2= 0 this is just a quadratic that can be solved ... so what do we come up for the other point?
yb1996
  • yb1996
wait, how do you solve for x in that equation? There's an x in the denominator, and I don't really know how the quadratic formula would work in that case.
amistre64
  • amistre64
the x is not in the denominator, but we can expand it out thru distribution and rewrok it into something like this x^2 + x/(2a) -(a^2+1/2) = 0 2a x^2 + x -(2a^3 + a) = 0 \[x=\frac{-1\pm\sqrt{1+4(2a^3+a)}}{4a}\]
yb1996
  • yb1996
so that would be the x value of point Q?
amistre64
  • amistre64
i wonder if we can simplify that any ... or if ive made an error in the process. as long as we can define the 2 points, we can find the distance between them
amistre64
  • amistre64
\[y=-\frac{1}{2a}(x-a)+a^2\] \[y=-\frac{x}{2a}+a^2+\frac12\] \[x^2+\frac{x}{2a}=a^2+\frac12\] x=a is obviously one solution
amistre64
  • amistre64
\[2ax^2+x-a(2a^2+1)=0\] \[x=\frac{-1\pm\sqrt{1+4a(2a^2+1)}}{4a}\] well,we have our 2 point regardless \[(x_1,x^2_1)~(x_2,x^2_2)\] both in terms of a \[d^2=(x_1-x_2)^2+(x^2_1-x^2_2 )^2\]
amistre64
  • amistre64
finding the derivatives with respect to a might be fun :)
amistre64
  • amistre64
the basic premise then takes us to : \[d^2 = n^2 + m^2\] \[2dd'=2nn'+2mm'\] \[d'=\frac{2nn'+2mm'}{2d}=0\] \[nn'+mm'=0\]
yb1996
  • yb1996
is d' dd/da, so you're taking the derivative with respect to a?
amistre64
  • amistre64
with respect to a, yes seeing how our distances rely upon x, when x=a ... the whole system depends on a i beleive
yb1996
  • yb1996
ok, so would the result that you got with implicit differentiation: nn' + mm' = 0 still work when you don't really have to differentiate implicitly in the regular problem?
amistre64
  • amistre64
yes, but the nm stuff was just generalizing it
yb1996
  • yb1996
because everything in the regular problem is in terms of a, so would that equate to: \[n \frac{ dn }{ da } + m \frac{ dm }{ da }\] ?
amistre64
  • amistre64
yes
amistre64
  • amistre64
spose n = x1 - x2 n = 1/2a^(-1) +4a^2 +2 n' = -1/2a^(-2) +8a
amistre64
  • amistre64
keep in mind that this is just a thought that is in my head for a solution, and there may exist simpler approaches :)
yb1996
  • yb1996
ok, and whatever value I get after I solve the equation should be the minimum distance between P and Q. right?
amistre64
  • amistre64
yes :) spose we let the points be defined as (a,a^2) and (f(a),[f(a)]^2) d^2 = (f(a)-a)^2 + ([f(a)]^2 - a^2)^2 2dd' = 2(f(a)-a)(f'(a)-a') + 2([f(a)]^2 - a^2)(2f(a) f'(a) - 2a) dd' = (f(a)-a) (f'(a)-a') + ([f(a)]^2 - a^2) (2f(a) f'(a) - 2a) d is irrelevant, or it may help to simplify when we divide it ... dunno, but when d'=0 we get 0 = (f(a)-a) (f'(a)-1) + ([f(a)]^2 - a^2) (2f(a) f'(a) - 2a) maybe we can solve this for a?and thereby determine the points p and q?
amistre64
  • amistre64
just another thought, may or not be worthwhile
yb1996
  • yb1996
ok, thank you so much more your time and your help! I really appreciate it!
amistre64
  • amistre64
youre welcome. good luck, and see what you can do to improve the method if you can.
yb1996
  • yb1996
Thanks!

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