# Calculating the tangential and normal vectors of an ellipse

1. Mar 5, 2016

### Nowayout

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. Mar 5, 2016

### geoffrey159

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
3. Mar 5, 2016

### Nowayout

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?

4. Mar 5, 2016

### geoffrey159

What is the determinant of two colinear vectors in the plane ?

5. Mar 5, 2016

Zero

6. Mar 5, 2016

### geoffrey159

Yes !
But now you are almost done so I can't help you anymore

7. Mar 5, 2016

### Nowayout

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?

8. Mar 5, 2016

### geoffrey159

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

9. Mar 5, 2016

### Nowayout

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...

10. Mar 5, 2016

### geoffrey159

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 $f'(t_0) = (-ay_0 / b, b x_0 /a )$.