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Find the points on the ellipse 100x^2+y^2=100 that are farthest away from the point (1,0)

Mathematics
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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)

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Other answers:

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? :\
\[d=\sqrt{(\Delta x)^2+(\Delta y)^2}=\sqrt{((x-1)^2+(y-0)^2}=\sqrt{(x-1)^2+y^2}\] I used the points given (1,0) and (x,y)
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
\[(\frac{1}{99}, \sqrt{100-100(\frac{1}{99})^2}) \text{ and } (\frac{1}{99},-\sqrt{100-100(\frac{1}{99})^2})\]
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. :)

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