Question

Find the point on the curve x=y^2-8y+ 18 that is closest to the point (-2, 4). Help?

Answer 1

x - 2 = (y-4)^2
parameterize this parabola

Answer 2

get it?

Answer 3

no

Answer 4

do u know wht a parabola is?

Answer 5

yes x^2

Answer 6

it can be anything of the form which has one variable of deg 1 and another of deg 2
like
(x-h) = (y-k)^2

Answer 7

okay so what does parameterize mean?

Answer 8

u write x and y in terms of a single variable t

Answer 9

d^2=(x-(-2))^2+(y-4)^2
d^2=(y^2-8y+18+2)^2+(y-4)^2
[d^2]'=2(y^2-8y+20)*(2y-8)+2(y-4)
set [d^2]'=0
(y^2-8y+20)(2y-8)+(y-4)=0
2(y^2-8y+20)(y-4)+(y-4)=0
(y-4){2(y^2-8y+20)+1}=0
(y-4){2y^2-16y+41}=0
y=4
y={16 + sqrt{16^2-4*2*41}}/4 -> this has imaginary part
y={16-sqrt{16^2-4*2*41}}/4-> this has imaginary part
so y=4
if y=4, then x=2
so the point (2,4) on the parabola is closest to (-2,4)

Answer 10

ya those are the steps ty soo much

Answer 11

im getting old

Answer 12

he was typing a reply for 10 min^^..knew it was gona be detailed

Answer 13

she

Answer 14

thank you him lol

Answer 15

you keep on showing ur smarter than me..how old r u?

Answer 16

26

Answer 17

fine then..to accept defeat..im 18

Answer 18

lol

Answer 19

you still have long way to go
you haven't started colllege yet right or first semester?

Answer 20

i still have long way to go

Answer 21

bt weve been taught a lot of this at school..in india..its jujst tht i dont think simple

Answer 22

what uni/college do you attend?

Answer 23

oh in india o.o

Answer 24

going to attend one

Answer 25

the list is yet to come out

Answer 26

I attend UALR
its a school in arkansas
im going to start my phd if i do well on the gre

Answer 27

phd on what?

Answer 28

congratz

Answer 29

im going the less math route (i mean it still has math involved, but its narrowed down to computational number theory) im going to do applied science but my researh will be of theory of cryptography

Answer 30

i figured out i'm not really a math person
i mean to be a math person you have to enjoy all math and i don't

Answer 31

too advanced for me

Answer 32

i still enjoy alot of math though
i mean i come on like everyday, but this stuff we do on here is mostly fun its computational its not really math lol

Answer 33

do i make sense lol

Answer 34

just ignore me lol

Answer 35

no ur making sense lol

Answer 36

so lazy did you understand everything i did up there?
i use the formula for distance squared

Answer 37

we wanted to minimize the distance so i took derivative and set =0

Answer 38

i understand the formula but not how you got the derivative in so little steps

Answer 39

i understand the formula but not how you got the derivative in so little steps

Answer 40

i understand the formula but not how you got the derivative in so little steps

Answer 41

i understand the formula but not how you got the derivative in so little steps

Answer 42

i used chain rule for both parts

Answer 43

everything is under a squre root so there should be a 1/2 at the start and the whole thing should be to the -1/2?

Answer 44

i see what you are saying
don't do the derivative of distance
take the derivative of distance squared
let f(y)=d^2
and find f'(y)

Answer 45

it is easier but lazy you can do derivative of d it will jjust take longer

Answer 46

hmm interesting , i never learned it this way

Answer 47

did you always do it unsquared?

Answer 48

yes

Answer 49

both ways are not wrong just so you know
you come to same conclusions both ways
minimizing d^2 will give you same result as minimizing d

Answer 50

how do i take the derivative your way?

Answer 51

so you have d^2 above right?
d^2=(y^2-8y+20)^2+(y-4)^2
f(y)=(y^2-8y+20)^2+(y-4)^2
f'(y)=2(y^2-8y+20)*(2y-8)+2(y-4)
used chain rule

Answer 52

yes

Answer 53

ty calculus exam tomorrow gota get some sleep. Night , appreciate the help

Answer 54

later

Answer 55

best of luck