Results 1 to 5 of 5

Math Help - Stuck on this problem

  1. #1
    Newbie
    Joined
    Jul 2008
    Posts
    10

    Stuck on this problem

    Find the points on the ellipse 4x^2+y^2=4 that are farthest away from the point (1,0).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member euclid2's Avatar
    Joined
    May 2008
    From
    Ottawa, Canada
    Posts
    400
    Awards
    1
    Quote Originally Posted by DroMan View Post
    Find the points on the ellipse 4x^2+y^2=4 that are farthest away from the point (1,0).
     4x^2 + y^2 = 4
     y = \pm \sqrt4-4x^2

    find the distance

    D: \sqrt (x-1)^2 + (y-1)^2
     \sqrt -3x^2 -2x + 5

    take the derivative and set it to 0.

     dd/dx = \frac{1}{2} (3x^2-2x+5) ^ (-1/2)(-6x-2)
     \frac{-6x-2}{2 \sqrt -3x^2 - 2x + 5} = 0

    multiply by the denom.

     -6x-2=0

     -6=2

     x= \frac {-1}{3}

    now plug into the ellipse and solve for y.
    Last edited by euclid2; November 23rd 2008 at 03:49 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,408
    Let d be the distance from the point (1,0) to the ellipse.

    Now if you remember your distance formula: d = \sqrt{(x - x_1)^2 + (y - y_1)^2} \ \Leftrightarrow \ d^2 = (x  - x_1)^2 + (y  - y_1)^2

    Here, (x_1, y_1) = (1,0) and (x , y ) is any arbitrary point on the ellipse.

    Now, arranging your equation of your ellipse we get: y^2 = 4-4x^2

    So, we want to find the absolute maximum of: d^2 = (x-1)^2 + y^2 \ \Leftrightarrow \ {\color{blue}d^2 = (x-1)^2 + (4-4x^2)}

    You do this by taking the derivative and setting it equal to 0 (note that taking the derivative of d^2 is equal to 2d d' by chain rule). Then determine if it is a maximum and you'll be done.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jul 2008
    Posts
    10
    Quote Originally Posted by euclid2 View Post
     4x^2 + y^2 = 4
     y = \pm \sqrt4-4x^2

    find the distance

    D: \sqrt (x-1)^2 + (y-1)^2
     \sqrt -3x^2 -2x + 5

    take the derivative and set it to 0.

     dd/dx = \frac{1}{2} (3x^2-2x+5) ^ (-1/2)(-6x-2)
     \frac{-6x-2}{2 \sqrt -3x^2 - 2x + 5} = 0

    multiply by the denom.

     -6x-2=0

     -6=2

     x= \frac {-1}{3}

    now plug into the ellipse and solve for y.
    how did you get the points for the distance formula? i do not understand what you did after, "multiply by the denom."
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member euclid2's Avatar
    Joined
    May 2008
    From
    Ottawa, Canada
    Posts
    400
    Awards
    1
    Quote Originally Posted by DroMan View Post
    how did you get the points for the distance formula? i do not understand what you did after, "multiply by the denom."
    Let me be more specific with the answer.
    So now d=√(√(4-4x)-0)+(x-1)) for the first half of the ellipse. simplify
    d=√(4-4x+x-2x+1)
    =√(-3x-2x+5)
    Now find derivative
    d'=(1/2)(-3x-2x+5)^(-1/2)(-6x-2)
    set equal to zero
    (1/2)(-3x-2x+5)^(-1/2)(-6x-2)
    which means
    -6x-2=0
    x=2/-6=-1/3
    Now we want maximum distance so we must test to see if this is a maximum.
    d'(-1)=1 <==positive
    d'(0)=-5^(-1/2) <==negative
    Therefore a maximum occurs at x=-1/3
    So now we find y value on ellipse
    4x+y=4
    4(-1/3)+y=4
    y=32/9
    y=√(32/9)=(4√2)/3
    Now even though we defined the ellipse in two parts it turns out that there's no need to attempt to find a maximum using the second half of the ellipse. Why? Take a look at this.
    The second half of the ellipse was defined asy2=-√(4-4x)
    So a point on the second half of the ellipse is defined as (x, -√(4-4x))
    The distance between a point on the second half of the ellipse and the point (1, 0) is d=√((-√(4-4x)-0)+(x-1))= √(4-4x+x-2x+1)]. Notice that this is the same thing we started with above? So we'd be optimizing the same thing again.
    Anyways, maximum distance is between[tex] (-1/3, (-4√2)/3) and (1, 0)and from(-1/3, (4√2)/3) to (1, 0)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. stuck on this problem for cal 1
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 16th 2008, 07:50 AM
  2. Stuck on a problem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 28th 2008, 12:36 PM
  3. here is another problem that i stuck on
    Posted in the Algebra Forum
    Replies: 2
    Last Post: June 28th 2007, 09:58 PM
  4. I got stuck on this problem: C(n,6)=C(n,9)
    Posted in the Algebra Forum
    Replies: 3
    Last Post: June 7th 2007, 08:22 AM
  5. Stuck on an SAT problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 30th 2006, 01:59 AM

Search Tags


/mathhelpforum @mathhelpforum