Question

Find the point closest..

Answer 1

http://prntscr.com/39tzhg

Answer 2

You have a directrix and a focus. They form a parabola. The vertex is the closest point to the directrix.

Answer 3

Where did parabola come from?

Answer 4

point + line = parabola.

Answer 5

I suppose you don't know what vectors are at this point?

Answer 6

I suppose we'll have to treat this like an optimization problem.

Answer 7

Let \(d\) be the distance.\[
d^2 = (x-(-4))^2+(y-1)^2 \implies d^2 = (x+4)^2 + (y-1)^2
\]Do you follow so far?

Answer 8

Used the distance formula.. used (-4, 1) and (x, y).. so on, so yea.

Answer 9

Now we differentiate with respect to \(x\): \[
2dd' = 2(x+4) + 2(y-1)y'
\]

Answer 10

We want to find critical points.

Answer 11

One thing to remember is that \(d\geq0\),

Answer 12

Another thing to remember is that: \[
5x+y=6
\]Differentiate this to get: \[
5+y'=0 \implies y'=-5
\]

Answer 13

Oh yeah, and \(y = 6-5x\).
Does this make sense so far?

Answer 14

So far so good.

Answer 15

Cause it gives us \[
2dd' = 2(x+4)+2(6-5x-1)(-5)
\]

Answer 16

Notice that if \(2dd'=0\), then that means \(d'=0\). Putting \(2d\) in the denominator doesn't eliminate any critical points.

Answer 17

So, can you simplify it and find the critical point? That is going to give you the x coordinate tat the minimum.

Answer 18

Since it is indeed a parabola, it will only have a minimum distance from the directrix, and it won't have any maximum.

Answer 19

@Luigi0210 Is that enough or do you need more help?

Answer 20

Again, let \(d'=0\) and solve for \(x\) to get the x coordinate of the closest point.

Answer 21

Yea, I think I got it from here, thanks :)

Answer 22

Minimise the distance between (-4,1) and (x,y) on the line which gives\[d^{2} = (x-(-4))^{2}+(y-1)^2=(x+4)^2+(y-1)^2\]but y=6-5x so substitute
\[d^{2} =(x+4)^2+(6-5x-1)^2=x^2+8x+16+25-50x+25x^2\]\[d^{2}=26x^2-42x+41\]Now a little trick. The x that gives the minimum value of d is going to be the same for x that gives the minimum d^2 so let y=d^2 and differentiate and equate to zero to get the x that gives the minimum value \[\frac{ d y }{ dx} = 52x-42 = 0, x=42/52=21/26\]Substitute x back in the line equation to get the y.

I'm sure there's a mistake along the way (I dont like 21/26) but I think you get the general idea.

Answer 23

Yup, thank you for your time too :)

Answer 24

another non-calc way is to solve the given line, and the perpendicular to it from given point :
http://www.wolframalpha.com/input/?i=5x%2By%3D6%2C+x-5y+%3D+-9

Answer 25

WTF LVL 99 AND ASK SUCH BIRTH-LEVEL QUESTION

Answer 26

A solution using Mathematica 9 is attached.

Answer 27

i seriously demand OS to reexamine the ability of people attributed as lvl 99

Answer 28

@johnz
Well I guess we all have to live with our mental limitations.

Answer 29

this question can be EASILY solved using vectors