Solution for tangent





      In this page 'Solution for tangent' we are going to see how to get the equations of tangents to the given ellipse in the page 'Example I'.

The problems are

  1. Find the equation of tangent to the ellipse 4x²+9y²=36 at (2,2).
  2. Find the equation of the tangent to the ellipse 3x²+2y²=5 at (-1,1).
  3. Find the equation of the tangent to the ellipse x²/16 + y²/12 =1 at (2,-3).  

Solution for tangent problem 1. 

            The given ellipse is  4x²+9y²=36.

                             4x²/36 +9y²/36  =1

                              x²/9  + y²/4     =  1

            Any  tangent to the given ellipse is of the form

                                     y = mx + √(a²m²+b²)

               Here a²=9 and b²=4

                                    y  =  mx + √(9m²+4)

              The tangent passes through (2,2)

                                  2  = m(2) +  √(9m²+4)

              Simplifying

                          2-2m     =   √(9m²+4)



                        (2-2m)²   =       9m² + 4

                  4+4m²-8m  =       9m² + 4

                     5m² +8m  =       0

                    m(5m+8) =       0

                                    m  =       0,  -8/5

             The two tangents are

                  y = (0)x + √(9(0)² +4)  and 

                                       y = (-8/5)x+√(9(-8/5)²+4)          

             Simplifying

                         y = 2  and y = (-8/5)x+26/5

                         y = 2  and 5y  = -8x+26 

                         y = 2 and  8x+5y = 26 are the required equations of tangents.

Solution for problem 2:

The equation of tangent at (x₁, y₁) is xx₁/a² + yy₁/b² = 1.

           The given equation is             3x²+2y² =5

                                          3x²/5 + 2y²/5    =1 

                                           3x²/5 + 2y²/5    =1

           Here (x₁, y₁) = (-1,1), a² =5/3 and b² = 5/2

           The required equation of tangent is

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

                                        -3x/5      +   2y/5      =  1

                                                  -3x+2y           =  5

                                                     -3x+2y-5     =  0

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

Solution for problem 3:

The equation of tangent at (x₁, y₁) is xx₁/a² + yy₁/b² = 1.

           The given equation is            x²/16+y²/12 =1

                                               x²/16 + y²/12    =1                  

          Here (x₁, y₁) = (2, -3), a² =16 and b² = 12

           The required equation of tangent is

                                      x(2)/16  +  y(-3)/12  = 1

                                         x/8    -    y/4      =  1

                                                  x-2y         =  8

                                                   x-2y-8     =  0

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


Problems for practice:

  1. Find the equation  of tangents to the ellipse from the point (0, 3) whose equation is 4x²+9y²=36.
  2. Find the equation of tangent at the point (1,1) to the ellipse
    2x²+3y² =5



           Parents and teachers can guide the students to solve the problems step by step as it is discussed in this page 'Solution for tangent'.

Students can try to solve the problems on their own, using the same methods discussed above. If you are having any doubt you can contact us through mail, we will help you to clear your doubts.





                                     Ellipse

                                      Home