anonymous
  • anonymous
Find the points on the ellipse 100x^2+y^2=100 that are farthest away from the point (1,0)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
freckles
  • freckles
You know the distance formula?
freckles
  • freckles
You want to maximize that using the points (1,0) and (x,y)
anonymous
  • anonymous
ok i know y^2=(100-100x^2)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
so d^2=x^2 + (100-100x^2)^2
anonymous
  • anonymous
but im not sure about the x^2 part is it (1-x)^2 because the point is (1,0)?
anonymous
  • anonymous
or am I just doing this all wrong? :\
freckles
  • freckles
\[d=\sqrt{(\Delta x)^2+(\Delta y)^2}=\sqrt{((x-1)^2+(y-0)^2}=\sqrt{(x-1)^2+y^2}\] I used the points given (1,0) and (x,y)
freckles
  • freckles
Okay and you squared the distance! Good job! :)
freckles
  • freckles
\[d^2=(x-1)^2+y^2\] Replace y^2 with what you found :)
anonymous
  • anonymous
yep i did that ! :o
freckles
  • freckles
\[100x^2+y^2=100 => y^2=100-100x^2\]
freckles
  • freckles
Why did you square y^2?
freckles
  • freckles
\[d^2=(x-1)^2+100-100x^2\]
anonymous
  • anonymous
OOOHH omg hahaahah thats where i went wrong?!
anonymous
  • anonymous
so then after i take the derivative is do i find the roots ? -199x-1=0? that doesnt seem right :s
freckles
  • freckles
Nope nope.... So when you found ((x-1)^2)' You used chain rule right?
freckles
  • freckles
We know that (100)'=0
freckles
  • freckles
(-100x^2)'=-100(x^2)' we know to use power rule here
anonymous
  • anonymous
yes so i get 2(x-1)*1 + 0 -200x
freckles
  • freckles
You need to distribute that 2. 2(x-1)=2x-1(2)=2x-2
anonymous
  • anonymous
okay!
freckles
  • freckles
Right!: )
freckles
  • freckles
\[\text{ Set } (d^2)'=0\]
freckles
  • freckles
\[2x-2-200x=0\]
anonymous
  • anonymous
ok but dont i have to divide by the 2d from when i differentiate (d^2)'
freckles
  • freckles
No no.
anonymous
  • anonymous
why not ?
freckles
  • freckles
Minimizing d^2 will give us the same results as minimizing d
anonymous
  • anonymous
ok, so i get x=1/98 ?
anonymous
  • anonymous
then would that be my point? x= 1/98 and subbing that into the original equation y= 9.99948?
anonymous
  • anonymous
also i only get 1 answer, shouldnt there be another point also?
freckles
  • freckles
\[y=\pm \sqrt{100-100x^2} \text{ correct?}\]
freckles
  • freckles
Say yes. lol
anonymous
  • anonymous
but my x is only 1 point? and my y's will be 2 points? sorry im just a little confused
freckles
  • freckles
\[(\frac{1}{99}, \sqrt{100-100(\frac{1}{99})^2}) \text{ and } (\frac{1}{99},-\sqrt{100-100(\frac{1}{99})^2})\]
anonymous
  • anonymous
okay but why 1/99? i thought we found that 2x-2-200x=0 then wouldnt x be 2/-198 ?
freckles
  • freckles
|dw:1353821740124:dw|
freckles
  • freckles
\[198/2=99\]
freckles
  • freckles
\[2/198=1/99\]
freckles
  • freckles
|dw:1353821859781:dw|
anonymous
  • anonymous
omg ok i see !!! i think i got it now !!!
freckles
  • freckles
|dw:1353821881622:dw|
anonymous
  • anonymous
thank you !! i get it now !!
freckles
  • freckles
Np. :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.