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.

Question

amistre64 · Answer

and what are your thoughts?

yb1996 · Answer

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 · Answer

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?

yb1996 · Answer

we take the derivative at that point

amistre64 · Answer

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 · Answer

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 · Answer

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 · Answer

we set that equation equal to x^2 ?

amistre64 · Answer

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 · Answer

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 · Answer

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 · Answer

so that would be the x value of point Q?

amistre64 · Answer

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 · Answer

$$y=-\frac{1}{2a}(x-a)+a^2$$
$$y=-\frac{x}{2a}+a^2+\frac12$$
$$x^2+\frac{x}{2a}=a^2+\frac12$$
x=a is obviously one solution

amistre64 · Answer

$$2ax^2+x-a(2a^2+1)=0$$
$$x=\frac{-1\pm\sqrt{1+4a(2a^2+1)}}{4a}$$
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 · Answer

finding the derivatives with respect to a might be fun :)

amistre64 · Answer

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 · Answer

is d' dd/da, so you're taking the derivative with respect to a?

amistre64 · Answer

with respect to a, yes  seeing how our distances rely upon x, when x=a ... the whole system depends on a i beleive

yb1996 · Answer

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 · Answer

yes, but the nm stuff was just generalizing it

yb1996 · Answer

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 · Answer

yes

amistre64 · Answer

spose n = x1 - x2

n = 1/2a^(-1) +4a^2 +2

n' = -1/2a^(-2) +8a

amistre64 · Answer

keep in mind that this is just a thought that is in my head for a solution, and there may exist simpler approaches :)

yb1996 · Answer

ok, and whatever value I get after I solve the equation should be the minimum distance between P and Q. right?

amistre64 · Answer

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 · Answer

just another thought, may or not be worthwhile

yb1996 · Answer

ok, thank you so much more your time and your help! I really appreciate it!

amistre64 · Answer

youre welcome.  good luck, and see what you can do to improve the method if you can.

yb1996 · Answer

Thanks!