Question

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

Answer 1

You know the distance formula?

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

or am I just doing this all wrong? :\

Answer 7

\[d=\sqrt{(\Delta x)^2+(\Delta y)^2}=\sqrt{((x-1)^2+(y-0)^2}=\sqrt{(x-1)^2+y^2}\]
I used the points given (1,0) and (x,y)

Answer 8

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

Answer 9

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

Answer 10

yep i did that ! :o

Answer 11

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

Answer 12

Why did you square y^2?

Answer 13

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

Answer 14

OOOHH omg hahaahah thats where i went wrong?!

Answer 15

so then after i take the derivative is do i find the roots ?

-199x-1=0? that doesnt seem right :s

Answer 16

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

Answer 17

We know that (100)'=0

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

okay!

Answer 22

Right!: )

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

No no.

Answer 27

why not ?

Answer 28

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

Answer 29

ok, so i get x=1/98 ?

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

Say yes. lol

Answer 34

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

Answer 35

\[(\frac{1}{99}, \sqrt{100-100(\frac{1}{99})^2}) \text{ and } (\frac{1}{99},-\sqrt{100-100(\frac{1}{99})^2})\]

Answer 36

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