Lesson Tangent lines and normal vectors to an ellipse

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Tangent lines and normal vectors to an ellipse

Tangent line to the ellipse  x%5E2%2Fa%5E2+%2B+y%5E2%2Fb%5E2+=+1  at the point  (x%5B0%5D,y%5B0%5D)  has the equation  x%2Ax%5B0%5D%2Fa%5E2%2By%2Ay%5B0%5D%2Fb%5E2=1.                                (1)
Tangent line to the ellipse  %28x-h%29%5E2%2Fa%5E2+%2B+%28y-k%29%5E2%2Fb%5E2+=+1  at the point  (x%5B0%5D,y%5B0%5D)  has the equation  %28x-h%29%2A%28x%5B0%5D-h%29%2Fa%5E2%2B%28y-k%29%2A%28y%5B0%5D-k%29%2Fb%5E2=1.       (2)
Let us prove the statement  (1)  now.  First, note that the straight line  x%2Ax%5B0%5D%2Fa%5E2%2By%2Ay%5B0%5D%2Fb%5E2=1  passes through the point  (x%5B0%5D,y%5B0%5D),  since  (x%5B0%5D,y%5B0%5D)  satisfies the equation  x%5E2%2Fa%5E2+%2B+y%5E2%2Fb%5E2+=+1.
Second, this straight line has only one common point with the ellipse  x%5E2%2Fa%5E2+%2B+y%5E2%2Fb%5E2+=+1.  Indeed, substitute the expression
y = %28b%5E2%2Fy%5B0%5D%29%2A%281-x%2Ax%5B0%5D%2Fa%5E2%29,  obtained from the straight line equation  x%2Ax%5B0%5D%2Fa%5E2%2By%2Ay%5B0%5D%2Fb%5E2=1,  into the ellipse equation  x%5E2%2Fa%5E2+%2B+y%5E2%2Fb%5E2+=+1.  You get

x%5E2%2Fa%5E2 + %28b%5E2%2Fy%5B0%5D%5E2%29%2A%281-x%2Ax%5B0%5D%2Fa%5E2%29%5E2 = 1.        (3)

Simplify the equation  (3)  step by step

a%5E2%2Ax%5E2+%2B+b%5E2%2Fy%5B0%5D%5E2%2A%28a%5E2-x%2Ax%5B0%5D%29%5E2+=+a%5E4,   a%5E2%2Ay%5B0%5D%5E2%2Ax%5E2+%2B+b%5E2%2A%28a%5E2-x%2Ax%5B0%5D%29%5E2+=+a%5E4%2Ay%5B0%5D%5E2,   ,   .

The discriminant of the last quadratic equation is  d = .

Now, note that  a%5E2%2Ay%5B0%5D%5E2%2Bb%5E2%2Ax%5B0%5D%5E2+=+a%5E2b%5E2  due to the ellipse equation  (1),  and  a%5E4%2Ab%5E2+-+a%5E4%2Ay%5B0%5D%5E2 = a%5E2%2A%28a%5E2%2Ab%5E2+-+a%5E2%2Ay%5B0%5D%5E2%29 = a%5E2b%5E2%2Ax%5B0%5D%5E2  due to the same reason.

Therefore, the discriminant  d  is equal to  d = 4a%5E4b%5E4%2Ax%5B0%5D%5E2+-+4a%5E2b%5E2%2Aa%5E2b%5E2%2Ax%5B0%5D%5E2 = 0.

This means that the equation  (3)  has only one root.  It is what has to be proved.

The proof of the statement  (2)  is similar to that of the statement  (1).
Or, you can introduce the new coordinate system  (x',y')  using translations  x' = x-h, y' = y-k.


Having these equations  (1)  and  (2)  for the tangent line we can make few steps further.

First, the straight line  x%2Ax%5B0%5D%2Fa%5E2%2By%2Ay%5B0%5D%2Fb%5E2=0  is parallel to the tangent line  x%2Ax%5B0%5D%2Fa%5E2%2By%2Ay%5B0%5D%2Fb%5E2=1  (because these two linear equations have no common solution)  and passes through the origin  (0,0)  of the coordinate system.

Second, the vectors (x,y) satisfying the equation  x%2Ax%5B0%5D%2Fa%5E2%2By%2Ay%5B0%5D%2Fb%5E2=0  are directed along this straight line.  Therefore, the vector  (x%5B0%5D%2Fa%5E2,y%5B0%5D%2Fb%5E2)  is the normal vector to the straight line  x%2Ax%5B0%5D%2Fa%5E2%2By%2Ay%5B0%5D%2Fb%5E2=0,  as the scalar product of the two vectors  (x,y)  and  (x%5B0%5D%2Fa%5E2,y%5B0%5D%2Fb%5E2)  is zero.  Hence, the vector  (x%5B0%5D%2Fa%5E2,y%5B0%5D%2Fb%5E2)  is the normal vector to the tangent line  x%2Ax%5B0%5D%2Fa%5E2%2By%2Ay%5B0%5D%2Fb%5E2=1,  too.

Finally, the vector  (-y%5B0%5D%2Fb%5E2,x%5B0%5D%2Fa%5E2)  is directed along the line  x%2Ax%5B0%5D%2Fa%5E2%2By%2Ay%5B0%5D%2Fb%5E2=0,  because this vector is orthogonal to the normal vector (x%5B0%5D%2Fa%5E2,y%5B0%5D%2Fb%5E2).  Hence, the vector (-y%5B0%5D%2Fb%5E2,x%5B0%5D%2Fa%5E2)  is directed along the tangent line to the ellipse  x%2Ax%5B0%5D%2Fa%5E2%2By%2Ay%5B0%5D%2Fb%5E2=1.

Thus, we proved that the vector  (-y%5B0%5D%2Fb%5E2,x%5B0%5D%2Fa%5E2)  is the tangent vector to the ellipse  (x%5B0%5D%5E2%2Fa%5E2+%2B+y%5B0%5D%5E2%2Fb%5E2+=+1)  at the point  (x%5B0%5D,y%5B0%5D),  while the vector  (x%5B0%5D%2Fa%5E2,y%5B0%5D%2Fb%5E2)  is the normal vector to the ellipse at this point.

If you want the  unit  tangent and normal vectors, you need to divide the two above vectors by their length, which is equal to  N = sqrt%28%28x%5B0%5D%2Fa%5E2%29%5E2+%2B+%28y%5B0%5D%2Fb%5E2%29%5E2%29.
So, the unit tangent vector and the unit normal vector are  (-y%5B0%5D%2F%28b%5E2%2AN%29,x%5B0%5D%2F%28a%5E2%2AN%29)  and  (x%5B0%5D%2F%28a%5E2%2AN%29,y%5B0%5D%2F%28b%5E2%2AN%29),  respectively.


Example 1.  Find the tangent line equation and the guiding vector of the tangent line to the ellipse  x%5E2%2F5%5E2+%2B+y%5E2%2F4%5E2+=+1  at the point  (3, 16%2F5).

First, let us check that the point  (3,16%2F5)  belongs to the ellipse  (Figure 1).  Indeed,
3%5E2%2F5%5E2+%2B+%281%2F4%5E2%29%2A%2816%2F5%29%5E2  =  9%2F25+%2B+%281%2F16%29%2A%28256%2F25%29  =  9%2F25+%2B+16%2F25  =  1.

According to the  statement 1  above, the equation of the tangent line to the ellipse at this point is        

x%2A3%2F5%5E2+%2B+%281%2F4%5E2%29%2A%28y%2A16%2F5%29+=+1,  or,  which is the same,  %283%2F25%29%2Ax+%2B+%281%2F5%29%2Ay+=+1.  The tangent line is shown in  Figure 1.



Figure 1.  To the  Example 1

The guiding vector of the tangent line is  (-1%2F5,3%2F25)  (anti-clockwise direction, shown in red in  Figure 1)  or  (1%2F5, -3%2F25)  (clockwise direction, shown in green in  Figure 1).

Summary

Tangent line to the ellipse  x%5E2%2Fa%5E2+%2B+y%5E2%2Fb%5E2+=+1  at the point  (x%5B0%5D,y%5B0%5D)  has the equation  x%2Ax%5B0%5D%2Fa%5E2%2By%2Ay%5B0%5D%2Fb%5E2=1.
Tangent line to the ellipse  %28x-h%29%5E2%2Fa%5E2+%2B+%28y-k%29%5E2%2Fb%5E2+=+1  at the point  (x%5B0%5D,y%5B0%5D)  has the equation  %28x-h%29%2A%28x%5B0%5D-h%29%2Fa%5E2%2B%28y-k%29%2A%28y%5B0%5D-k%29%2Fb%5E2=1.
The normal vector to the ellipse  x%5E2%2Fa%5E2+%2B+y%5E2%2Fb%5E2+=+1  at the point  (x%5B0%5D,y%5B0%5D)  is  (x%5B0%5D%2Fa%5E2, y%5B0%5D%2Fb%5E2)  (outward)  or  (-x%5B0%5D%2Fa%5E2, -y%5B0%5D%2Fb%5E2)  (inward).
The unit normal vector to the ellipse  x%5E2%2Fa%5E2+%2B+y%5E2%2Fb%5E2+=+1  at the point  (x%5B0%5D,y%5B0%5D)  is  (x%5B0%5D%2F%28a%5E2%2AN%29,y%5B0%5D%2F%28b%5E2%2AN%29)  (outward)  or  (-x%5B0%5D%2F%28a%5E2%2AN%29,-y%5B0%5D%2F%28b%5E2%2AN%29)  (inward) ,
    where N = sqrt%28x%5B0%5D%5E2%2Fa%5E4+%2B+y%5B0%5D%5E2%2Fb%5E4%29.
The guiding vector of the tangent line to the ellipse  x%5E2%2Fa%5E2+%2B+y%5E2%2Fb%5E2+=+1  at the point  (x%5B0%5D,y%5B0%5D) is  (-y%5B0%5D%2Fb%5E2,x%5B0%5D%2Fa%5E2)  (anti-clockwise direction)  or
    (y%5B0%5D%2Fb%5E2,-x%5B0%5D%2Fa%5E2)  (clockwise direction).
The unit guiding vector of the tangent line to the ellipse  x%5E2%2Fa%5E2+%2B+y%5E2%2Fb%5E2+=+1  at the point  (x%5B0%5D,y%5B0%5D)  is  (-y%5B0%5D%2F%28b%5E2%2AN%29,x%5B0%5D%2F%28a%5E2%2AN%29)  (anti-clockwise direction)  or
    (y%5B0%5D%2F%28b%5E2%2AN%29,-x%5B0%5D%2F%28a%5E2%2AN%29)  (clockwise direction),  where N = sqrt%28x%5B0%5D%5E2%2Fa%5E4+%2B+y%5B0%5D%5E2%2Fb%5E4%29.


The formulas found in this lesson are used in the next lesson  Optical property of an ellipse  to prove the  optical property of an ellipse.

My other lessons on ellipses in this site are
    - Ellipse definition, canonical equation, characteristic points and elements
    - Ellipse focal property
    - Tangent lines and normal vectors to a circle
    - Optical property of an ellipse
    - Optical property of an ellipse revisited

    - Standard equation of an ellipse
    - Identify elements of an ellipse given by its standard equation
    - Find the standard equation of an ellipse given by its elements

    - General equation of an ellipse
    - Transform a general equation of an ellipse to the standard form by completing the square
    - Identify elements of an ellipse given by its general equation

    - Standard equation of a circle
    - Find the standard equation of a circle
    - General equation of a circle
    - Transform general equation of a circle to the standard form by completing the squares
    - Identify elements of a circle given by its general equation

    - OVERVIEW of lessons on ellipses

Use this file/link  ALGEBRA-II - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-II.

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