Question

Consider the parabola y = x^2. Find the point on the parabola that is closest to the point (3, 1).

Answer 1

this is the problem of finding the minima

closest to point, means MINIMUM distance from that point

Answer 2

let the point be (x,y)

distance between x,y and 3,1
will be given by ?

Answer 3

\[1=3^2\]

Answer 4

??

Answer 5

use the distance formula
Distance between points (x1,y1) and (x2,y2) is
\(\huge d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)

Answer 6

uuu.....I'll have to put it in that order right?

Answer 7

order doesn't matter
x1 - x2 or x2 - x1
both are same

Answer 8

here,
x1 = x
y1 = y = x^2

x2 = 3
y2 = 1

Answer 9

So..what's next?

Answer 10

plug those in the distance formula

Answer 11

Ok...Give me a sec:)

Answer 12

\[d=\sqrt{(x-3)^2}+(x^2-1)\]

Answer 13

Right??

Answer 14

to simplify things, we can also minimize the square of distance = d^2
so that we could ignore the square root sign,

and your function just becomes
\(f(x) = (x-3)^2+ (x^2-1)^2\)

Answer 15

So, that's my final answer..right?

Answer 16

did you cancel out
square root and square for x^2-1 ??

\(\sqrt{a^2+b^2} \ne \sqrt {(a^2)} +b \)

Answer 17

Yes. I did!:)

Answer 18

that was not correct thing to do ....

and do you know how to minimize the function ?

Answer 19

My bad.. and no..I don't know..

Answer 20

whats the topic of this question ?

derivatives ?

Answer 21

What? I don't know what exactly you're asking for...

Answer 22

pre-calc, right ?
any sub-topic?
any similar solved question ?
so that i know which method to use to explain...

Answer 23

Hang on..I'm loosing connection real bad..As you can see....I'm replying late...so I'll just get back a little later.When it's better...

Answer 24

no problem :)

Answer 25

Ok. @hartnn Yes. It is Pre-calculus. No subtopic and I don't have any similar solved question yet..This is my first one...

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Answer 41

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