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You know the distance formula?

You want to maximize that using the points (1,0) and (x,y)

ok i know y^2=(100-100x^2)

so d^2=x^2 + (100-100x^2)^2

but im not sure about the x^2 part is it (1-x)^2 because the point is (1,0)?

or am I just doing this all wrong? :\

Okay and you squared the distance!
Good job! :)

\[d^2=(x-1)^2+y^2\]
Replace y^2 with what you found :)

yep i did that ! :o

\[100x^2+y^2=100 => y^2=100-100x^2\]

Why did you square y^2?

\[d^2=(x-1)^2+100-100x^2\]

OOOHH omg hahaahah thats where i went wrong?!

so then after i take the derivative is do i find the roots ?
-199x-1=0? that doesnt seem right :s

Nope nope....
So when you found ((x-1)^2)'
You used chain rule right?

We know that (100)'=0

(-100x^2)'=-100(x^2)'
we know to use power rule here

yes so i get 2(x-1)*1 + 0 -200x

You need to distribute that 2.
2(x-1)=2x-1(2)=2x-2

okay!

Right!: )

\[\text{ Set } (d^2)'=0\]

\[2x-2-200x=0\]

ok but dont i have to divide by the 2d from when i differentiate (d^2)'

No no.

why not ?

Minimizing d^2 will give us the same results as minimizing d

ok, so i get x=1/98 ?

then would that be my point? x= 1/98 and subbing that into the original equation y= 9.99948?

also i only get 1 answer, shouldnt there be another point also?

\[y=\pm \sqrt{100-100x^2} \text{ correct?}\]

Say yes. lol

but my x is only 1 point? and my y's will be 2 points? sorry im just a little confused

okay but why 1/99? i thought we found that 2x-2-200x=0
then wouldnt x be 2/-198 ?

|dw:1353821740124:dw|

\[198/2=99\]

\[2/198=1/99\]

|dw:1353821859781:dw|

omg ok i see !!! i think i got it now !!!

|dw:1353821881622:dw|

thank you !! i get it now !!

Np. :)