Example II  

                   In this page 'Example II' we are going to see how to get the equation of normal for the given ellipse. 

Example 1: 

            Find the equation of normal to the ellipse x²/4 + y²/12 =1 at (-1,3).

 Solution:

            The equation of normal at (x₁, y₁) is

                                  a²x/x₁ - b²y/y₁ = a²-b².

            Here (x₁, y₁) is (-1, 3), a² =4 and b² =12

            The required equation of normal is

                                 4x/(-1) - 12y/3  =   4-12

            Simplifying,

                                  -4x     - 4y       =     -8

                                            x + y     =  2

            The above equation is the required equation of normal to the given ellipse.

Example II: 

Find the equation of the normal to the ellipse x²/16+y²/48 = 1 at (2,4).

Solution:

            The equation of normal at (x₁, y₁) is

                                  a²x/x₁ - b²y/y₁ = a²-b².

            Here (x₁, y₁) is (2,4), a² =16 and b² =48

            The required equation of normal is

                                 16x/(2) - 48y/4  =   16-48

            Simplifying,

                                    8x     - 12y    =     -32

                                            2x -3y   =  -8

            The above equation is the required equation of normal to the given ellipse.

Example 3: 

Find the equation of the normal to the ellipse 3x²+2y²=5 at (-1,1).

Solution: 

               The equation of normal at (x₁, y₁) is

                                  a²x/x₁ - b²y/y₁ = a²-b². 

               The equation of normal at (x₁, y₁) is

                                  x²/5/3 - y²/5/2 = 1.

                         5/3 (x)/(-1) - 5/2(y)/1 =  5/3 - 5/2

                            5/3 (x) +  5/2 (y)     =   5/6

                                 10x + 15y          =    5

                                     2x + 3y     =  1

          The above equation is the required equation of the normal.

Example 4: 

   The normal at the end of the latus rectum of the ellipse x²/a² +y²/b² =1 intersects the major axis at G. Prove that CG = ae³ where C is the centre of the ellipse.

Solution:

   Latus rectum of the ellipse at (ae, b²/a)

   The equation of the normal drawn is   a²x/ae - b²ya/b² = a²-b²

                                                             ax/e - ay         = a²-b²

   This meets 'a' on the major axis, y=0

                              ax/e               = a²-b² =e²a²

                                            x      = e²a².e/a

                                            x      =  ae³



Practice problems:

1. Find the equation of normal to the ellipse x²/36+y²/49 = 1 at (2,4).
2. Find the equation of normal to the ellipse 2x²+3y²=5 at (1,-1).
3. Find the equation of normal to the ellipse 4x²+9y²=36 at (2,2).

Parents and teachers can guide the students to go through the examples discussed in page 'Example I' step by step. Students can solve the problems using the same method discussed above. If you are having any doubt you can contact us through mail, we will help you to clear your doubts.

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