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Calculating the tangential and normal vectors of an ellipse

  1. Mar 5, 2016 #1
    1. The problem statement, all variables and given/known data

    The ellipse is given as (x^2/a^2) + (y^2/b^2)=1

    I´m meant to calculate a tangential vector, a normal vector and find an equation for the tangent using a random point (x0,y0).

    I´m meant to do this in 2 ways: firstly by using the parametrization x(t)=a*cos(t) and y(t)=b*sin(t) and secondly by solving the equation for x or y.


    2. Relevant equations

    The formula for a tangential vector is ((dx/dt) , (dy/dt)).

    The formula for a normal vector is +- ((dx/dt) , (dy/dt))/||((dx/dt) , (dy/dt))||

    The tangent is linear, so it should have the form y=kx+d


    3. The attempt at a solution

    x(t)=a*cos(t) and y(t)=b*sin(t), so the tangential vector is ((dx/dt) , (dy/dt)) = (-a*sin(t), b*cos(t)) = (-ay/b , bx/a)

    ||((dx/dt) , (dy/dt))|| = \sqrt{(ay/b)^2 + (bx/a)^2} , so the normal vector is (-ay/b , bx/a)/(\sqrt{(ay/b)^2 + (bx/a)^2})

    I´m not sure how to find an equation for the tangent. I can use y=kx+d and I think that k is (-(a*y0)/b , (b*x0)/a) , but I don´t know how to continue from there.



    For the second part:

    x = \sqrt{a^2- (y^2*a^2)/b^2}

    y= \sqrt{b^2-(x^2*b^2)/a^2}

    So do I just have to calculate (dx/dy, dy/dx) now?
     
  2. jcsd
  3. Mar 5, 2016 #2
    Yes, your ellipse is parametrized by ##f(t)=(a\cos(t), b\cos(t) )##.
    A tangent vector at point ##M(t) = f(t)## to the ellipse is given by the derivative of arc ##f## at ##t##.

    The tangent line to the ellipse at point ##M_0(x_0,y_0)## is the set of points ##M(x,y)## such that ##\vec{M_0M}## belongs to the vector line directed by the tangent vector at ##M_0## (EDIT: in other terms, if ##M_0 = f(t_0)##, then ##\vec{M_0M}## and ##f'(t_0)## are colinear). How would you turn this into a cartesian equation ?
     
    Last edited: Mar 5, 2016
  4. Mar 5, 2016 #3
    Thank you. So the tangent and normal vectors are correct?

    How do I work out the vector line directed by the tangent vector at ##M_0## ? The tangent vector is (-ay/b , bx/a), how do I work out a line from that?
     
  5. Mar 5, 2016 #4
    What is the determinant of two colinear vectors in the plane ?
     
  6. Mar 5, 2016 #5
    Zero
     
  7. Mar 5, 2016 #6
    Yes !
    But now you are almost done so I can't help you anymore
     
  8. Mar 5, 2016 #7
    if ##M_0=f(t_0)## , then ##{M_0M}## and ##f'(t_0)## are collinear, so the determinant must be zero. That means that
    det ( ## \begin{array}{ccc}
    t-x_0 & -(a*y_0)/b \\
    t-y_0 & (b*x_0)/a
    \end{array} ## ) = (t-x_0)*(b*x_0)/a + (a*y_0)/b * t-y_0 =0


    Am I on the right track?
     
  9. Mar 5, 2016 #8
    No, ##M(x,y)## is not equatl to ##(t,t)## !
    The answer is ##\frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1 ##, but you must find it yourself
     
  10. Mar 5, 2016 #9
    I´m sorry, I´m stuck here.

    ##M(x,y) = ( \begin{array} {c} a*cos(t) \\ b*sin(t) \end{array} )##

    ##M_0## is what I´m trying to work out.

    So the determinant of ##M_0M## and ##f'(t_0)## is ##det ( \begin{array} {cc} a*cos(t)-M_0 & -(a*y_0)/b \\ b*sin(t)-M_0 & (b*x_0)/a \end{array} ) = (a*cos(t)-M_0)*(b*x_0)/a + (b*sin(t)-M_0)*(a*y_0)/b = 0 ## , but I don´t think that just using ##M_0## is correct...
     
  11. Mar 5, 2016 #10
    You are looking for a cartesian equation of the tangent, meaning that you must give up the parametric representation by eliminating the parameter. You are left with :
    ##\vec{M_0M} = (x-x_0, y-y_0) ## and ##f'(t_0) = (-ay_0 / b, b x_0 /a ) ##.
     
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